1061: For Hilbert Space, Nonempty Closed Convex Subset, and Point on Hilbert Space, There Is Unique Point on Subset Whose Distance to Point Is Minimum

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description/proof of that for Hilbert space, nonempty closed convex subset, and point on Hilbert space, there is unique point on subset whose distance to point is minimum

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Topics

About: vectors space
About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Hilbert space.
• The reader knows a definition of closed set.
• The reader admits the parallelogram law on any vectors space normed induced by any inner product.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
• The reader admits the proposition that any metric is continuous with respect to the topology induced by the metric.
• The reader admits the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
• The reader admits the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.

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Target Context

• The reader will have a description and a proof of the proposition that for any Hilbert space, any nonempty closed convex subset, and any point on the Hilbert space, there is the unique point on the subset whose distance to the point is the minimum.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V$: $\in \{\text{ the }F\text{ Hilbert spaces }\}$, with $⟨\bullet ,\bullet ⟩:V\times V\to F$, $\Vert \bullet \Vert :V\to \mathbb{R}$, and $dist:V\times V\to \mathbb{R}$
$C$: $\in \{\text{ the nonempty closed convex subsets of }V\}$
$w$: $\in V$
//

Statements:
$!\mathrm{\exists }{v}_0\in C(dist(w,v_0)=inf\{dist(w,v)|v\in C\})$
//

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2: Note

The space's being Hilbert is crucial.

For example, when the space is the metric topological subspace of the Euclidean metric space, ${\mathbb{R}}^2$, that (the subspace) is the union of the open unit disc and the point, $(0,-2)$, the open unit disc is a nonempty closed convex subset (the intersection of the closed unit disk on ${\mathbb{R}}^2$ and the subspace), and the infimum of the distances from $(0,-2)$ is $1$, but there is no point on the closed subset whose distance to the point is $1$, in fact, there is no point on the space whose distance to the point is $1$.

Also the closed subset's being convex is crucial for Proof: at least, obviously, the uniqueness does not generally hold if the closed subset is not convex.

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3: Proof

Whole Strategy: Step 1: see that infimum of the distances, $d$, exists; Step 2: take any sequence of points on $C$ whose distances converge to $d$; Step 3: see that the sequence is a Cauchy sequence; Step 4: see that the convergence is on $C$; Step 5: see that the distance of the convergence to $w$ is $d$; Step 6: see that there is no other point on $C$ whose distance to $w$ is $d$.

Step 1:

As $\{dist(w,v)|v\in C\}$ is a subset of $\mathbb{R}$ lower bounded by $0$, $\{dist(w,v)|v\in C\}$ has the infimum $d\in \mathbb{R}$: which is a nature of $\mathbb{R}$.

Step 2:

Let us take any sequence of points on $C$, $v_1,v_2,...$, such that $d\le dist(w,v_j)<d+1/2^j$, which is possible because $d=inf\{dist(w,v)|v\in C\}$: there may be some duplications, which does not matter.

$dist(w,v_1),dist(w,v_2),...$ converges to $d$.

Step 3:

Let us see that the sequence of points is a Cauchy sequence.

$dist(v_m,v_n)=\Vert v_n-v_m\Vert =\Vert (v_n-w)-(v_m-w)\Vert$.

By the parallelogram law on any vectors space normed induced by any inner product, $\Vert (v_n-w)-(v_m-w){\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-\Vert (v_n-w)+(v_m-w){\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-\Vert v_n+v_m-2w{\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-4\Vert (v_n+v_m)/2-w{\Vert }^2$.

For any $0<ϵ$, there is an $N\in \mathbb{N}$ such that for each $N<m,n$, $\Vert v_m-w{\Vert }^2,\Vert v_n-w{\Vert }^2<d^2+{ϵ}^2/4$.

$(v_n+v_m)/2\in C$, because $C$ is convex, so, $d\le \Vert (v_n+v_m)/2-w\Vert$.

So, $\Vert v_n-v_m{\Vert }^2=2(\Vert v_n-w{\Vert }^2+\Vert v_m-w{\Vert }^2)-4\Vert (v_n+v_m)/2-w{\Vert }^2<2(2(d^2+{ϵ}^2/4))-4d^2={ϵ}^2$.

So, the sequence of points is a Cauchy sequence.

Step 4:

As $V$ is complete, there is the convergence of the sequence, $v_0\in V$.

$v_0$ is an accumulation point of $C$, because for each open ball around $v_0$, there is a $v_j\in C$ contained in the open ball, because the sequence converges to $v_0$.

By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, $v_0$ is on the closure of $C$, but the closure of $C$ is $C$, because $C$ is closed.

So, $v_0\in C$.

Step 5:

The distance map, $dist(w,v):V\to \mathbb{R}$, with $w$ fixed is a continuous map, by the proposition that any metric is continuous with respect to the topology induced by the metric and the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.

As $dist(w,v_1),dist(w,v_2),...$ converges to $d$, $dist(w,v_0)=d$, by the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.

So, there is at least 1 $v_0\in C$ such that $dist(w,v_0)=inf\{dist(w,v)|v\in C\}$.

Step 6:

Let us see that there is no other point on $C$ whose distance to $w$ is $d$.

Let us suppose there was another $v_0^{\prime }\in C$ such that $dist(w,v_0^{\prime })=d$.

$\Vert v_0-v_0^{\prime }\Vert =\Vert (v_0-w)-(v_0^{\prime }-w)\Vert$.

By the parallelogram law on any vectors space normed induced by any inner product, $\Vert (v_0-w)-(v_0^{\prime }-w){\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-\Vert (v_0-w)+(v_0^{\prime }-w){\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-\Vert (v_0+v_0^{\prime })-2w{\Vert }^2=2(\Vert v_0-w{\Vert }^2+\Vert v_0^{\prime }-w{\Vert }^2)-4\Vert (v_0+v_0^{\prime })/2-w{\Vert }^2$, but $\Vert v_0-w\Vert =dist(w,v_0)=d$, $\Vert v_0^{\prime }-w\Vert =dist(w,v_0^{\prime })=d$, and as $(v_0+v_0^{\prime })/2\in C$, $d\le \Vert (v_0+v_0^{\prime })/2-w\Vert$, so, $\le 2(2d^2)-4d^2=0$, which implies that $\Vert (v_0-w)-(v_0^{\prime }-w){\Vert }^2=0$, which implies that $v_0=v_0^{\prime }$.

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References

<The previous article in this series | The table of contents of this series |

1060: For Continuous Map and Net with Directed Index Set That Converges to Point on Domain, Image of Net Converges to Image of Point and if Codomain Is Hausdorff, Convergence of Image of Net Is Image of Point

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description/proof of that for continuous map and net with directed index set that converges to point on domain, image of net converges to image of point and if codomain is Hausdorff, convergence of image of net is image of point

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of convergence of net with directed index set.
• The reader knows a definition of continuous map.
• The reader admits the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$D$: $\in \{\text{ the directed index sets }\}$
$N$: $:D\to T_1$, $\in \{\text{ the nets with }D\}$
$t$: $\in T_1$, $\in \{\text{ the convergences of }N\}$
$f(t)$: $\in T_2$
//

Statements:
$f(t)\in \{\text{ the convergences of }f\circ N\}$
$\wedge$
(
$T_2\in \{\text{ the Hausdorff topological spaces }\}$
$⟹$
$limf\circ N=f(t)$
)
//

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2: Note

"the convergence of net" makes sense only when there is the unique convergence of the net.

So, 'the net converges to a point' is different from 'the convergence of the net is the point': the latter means that there is the unique convergence of the net.

The 1st-half of this proposition says that $f\circ N$ converges to $f(t)$; the 2nd-half of this proposition says that $f\circ N$ has the unique convergence and the convergence is $f(t)$.

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3: Proof

Whole Strategy: Step 1: see that $f\circ N$ is a net with $D$; Step 2: take any neighborhood of $f(t)$, $N_{f(t)}$; Step 3: take an open neighborhood of $t$, $U_t$, such that $f(U_t)\subseteq N_{f(t)}$; Step 4: take a $j_0\in D$ such that for each $j\in D$ such that $j_0\le j$, $N(j)\in U_t$; Step 5: see that $f\circ N(j)\in N_{f(t)}$; Step 6: suppose that $T_2$ is Hausdorff, and see that the convergence of $f\circ N$ is unique and is $f(t)$.

Step 1:

$f\circ N$ is $:D\to T_1\to T_2$, and so, is a net with $D$.

Step 2:

Let us take any neighborhood of $f(t)$, $N_{f(t)}\subseteq T_2$.

Step 3:

As $f$ is continuous, there is an open neighborhood of $t$, $U_t\subseteq T_1$, such that $f(U_t)\subseteq N_{f(t)}$.

Step 4:

As $N$ converges to $t$, there is a $j_0\in D$ such that for each $j\in D$ such that $j_0\le j$, $N(j)\in U_t$.

Step 5:

For each $j\in D$ such that $j_0\le j$, $f\circ N(j)\in f(U_t)\subseteq N_{f(t)}$, which means that $f\circ N$ converges to $f(t)$.

But that does not necessarily mean that $f(t)$ is the unique convergence. So, it is not warranted to talk about "the convergence of $f\circ N$".

Step 6:

Let us suppose that $T_2$ is Hausdorff.

By the proposition that for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most, there is at most 1 convergence, but there is indeed a convergence, $f(t)$, and so, $f(t)$ is the unique convergence of $f\circ N$.

So, it is warranted to denote $limf\circ N=f(t)$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1059: Metric Is Continuous w.r.t. Topology Induced by Metric

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description/proof of that metric is continuous w.r.t. topology induced by metric

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of metric.
• The reader knows a definition of topology induced by metric.
• The reader knows a definition of product topology.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that any metric is continuous with respect to the topology induced by the metric.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the metric spaces }\}$, with metric, $dist:T\times T\to \mathbb{R}$ with the topology induced by $dist$
$\mathbb{R}$: $=\text{ the Euclidean topological space }$
//

Statements:
$dist\in \{\text{ the continuous maps }\}$
//

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2: Proof

Whole Strategy: Step 1: take any $p=(p^1,p^2)\in T\times T$, any neighborhood of $dist(p^1,p^2)$, $N_{dist(p^1,p^2)}\subseteq \mathbb{R}$, and an open ball around $dist(p^1,p^2)$, $B_{dist(p^1,p^2),ϵ}\subseteq \mathbb{R}$, such that $B_{dist(p^1,p^2),ϵ}\subseteq N_{dist(p^1,p^2)}$; Step 2: take the open neighborhood of $p$, $B_{p^1,ϵ/2}\times B_{p^2,ϵ/2}\subseteq T\times T$, and see that $dist(B_{p^1,ϵ/2}\times B_{p^2,ϵ/2})\subseteq B_{dist(p^1,p^2),ϵ}\subseteq N_{dist(p^1,p^2)}$.

Step 1:

Let $p=(p^1,p^2)\in T\times T$ be any.

Let $N_{dist(p^1,p^2)}\subseteq \mathbb{R}$ be any neighborhood of $dist(p^1,p^2)$.

There is an open ball around $dist(p^1,p^2)$, $B_{dist(p^1,p^2),ϵ}\subseteq \mathbb{R}$, such that $B_{dist(p^1,p^2),ϵ}\subseteq N_{dist(p^1,p^2)}$.

Step 2:

Let us take the open neighborhood of $p$, $B_{p^1,ϵ/2}\times B_{p^2,ϵ/2}\subseteq T\times T$, which is indeed open on $T\times T$ by the definition of topology induced by metric and the definition of product topology.

For any point, $p^{\prime }=(p^{\prime 1},p^{\prime 2})\in B_{p^1,ϵ/2}\times B_{p^2,ϵ/2}$, $dist(p^{\prime 1},p^{\prime 2})\le dist(p^{\prime 1},p^1)+dist(p^1,p^{\prime 2})\le dist(p^{\prime 1},p^1)+dist(p^1,p^2)+dist(p^2,p^{\prime 2})<ϵ/2+dist(p^1,p^2)+ϵ/2$, so, $dist(p^{\prime 1},p^{\prime 2})-dist(p^1,p^2)<ϵ$; $dist(p^1,p^2)\le dist(p^1,p^{\prime 1})+dist(p^{\prime 1},p^2)\le dist(p^1,p^{\prime 1})+dist(p^{\prime 1},p^{\prime 2})+dist(p^{\prime 2},p^2)<ϵ/2+dist(p^{\prime 1},p^{\prime 2})+ϵ/2$, so, $dist(p^1,p^2)-dist(p^{\prime 1},p^{\prime 2})<ϵ$; so, $|dist(p^{\prime 1},p^{\prime 2})-dist(p^1,p^2)|<ϵ$.

That means that $dist(B_{p^1,ϵ/2}\times B_{p^2,ϵ/2})\subseteq B_{dist(p^1,p^2),ϵ}\subseteq N_{dist(p^1,p^2)}$. So, $dist$ is continuous.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1058: Parallelogram Law on Vectors Space Normed Induced by Inner Product

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description/proof of parallelogram law on vectors space normed induced by inner product

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of norm induced by inner product on real or complex vectors space.

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Target Context

• The reader will have a description and a proof of the parallelogram law on any vectors space normed induced by any inner product.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$, with $\Vert \bullet \Vert :V\to \mathbb{R}$ induced by $⟨\bullet ,\bullet ⟩:V\times V\to F$
//

Statements:
$\mathrm{\forall }{v}_1,v_2\in V(\Vert v_1+v_2{\Vert }^2+\Vert v_1-v_2{\Vert }^2=2(\Vert v_1{\Vert }^2+\Vert v_2{\Vert }^2))$
//

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2: Proof

Whole Strategy: Step 1: expand $\Vert v_1+v_2{\Vert }^2=⟨v_1+v_2,v_1+v_2⟩$ and $\Vert v_1-v_2{\Vert }^2=⟨v_1-v_2,v_1-v_2⟩$, and see that the sum is $2(⟨v_1,v_1⟩+⟨v_1,v_1⟩)=2(\Vert v_1{\Vert }^2+\Vert v_2{\Vert }^2)$.

Step 1:

$\Vert v_1+v_2{\Vert }^2=⟨v_1+v_2,v_1+v_2⟩=⟨v_1,v_1⟩+⟨v_2,v_2⟩+⟨v_1,v_2⟩+⟨v_2,v_1⟩$.

$\Vert v_1-v_2{\Vert }^2=⟨v_1-v_2,v_1-v_2⟩=⟨v_1,v_1⟩+⟨v_2,v_2⟩-⟨v_1,v_2⟩-⟨v_2,v_1⟩$.

So, $\Vert v_1+v_2{\Vert }^2+\Vert v_1-v_2{\Vert }^2=2⟨v_1,v_1⟩+2⟨v_2,v_2⟩=2(⟨v_1,v_1⟩+⟨v_2,v_2⟩)=2(\Vert v_1{\Vert }^2+\Vert v_2{\Vert }^2)$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1057: Topological Space Induced by Metric Is Hausdorff

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that topological space induced by metric is Hausdorff

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of topology induced by metric.
• The reader knows a definition of Hausdorff topological space.

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Target Context

• The reader will have a description and a proof of the proposition that the topological space induced by any metric is Hausdorff.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$, induced by any metric, $dist:T\times T\to \mathbb{R}$
//

Statements:
$T\in \{\text{ the Hausdorff topological spaces }\}$
//

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2: Proof

Whole Strategy: Step 1: for each $t,t^{\prime }\in T$ such that $t\ne t^{\prime }$, take the open neighborhood of $t$, $B_{t,dist(t,t^{\prime })/2}$, and the open neighborhood of $t^{\prime }$, $B_{t^{\prime },dist(t,t^{\prime })/2}$,, and see that $B_{t,dist(t,t^{\prime })/2}\cap B_{t^{\prime },dist(t,t^{\prime })/2}=\mathrm{\varnothing }$.

Step 1:

Let $t,t^{\prime }\in T$ be any such that $t\ne t^{\prime }$.

$0<dist(t,t^{\prime })$.

Let us take the open neighborhood of $t$, $B_{t,dist(t,t^{\prime })/2}$, and the open neighborhood of $t^{\prime }$, $B_{t^{\prime },dist(t,t^{\prime })/2}$,.

Let us see that $B_{t,dist(t,t^{\prime })/2}\cap B_{t^{\prime },dist(t,t^{\prime })/2}=\mathrm{\varnothing }$.

Let $u\in B_{t,dist(t,t^{\prime })/2}$ be any.

$dist(t,t^{\prime })\le dist(t,u)+dist(u,t^{\prime })$, so, $dist(t,t^{\prime })-dist(t,u)\le dist(u,t^{\prime })$, but $dist(t,t^{\prime })/2=dist(t,t^{\prime })-dist(t,t^{\prime })/2<dist(t,t^{\prime })-dist(t,u)$, so, $dist(t,t^{\prime })/2<dist(u,t^{\prime })$, which means that $u\notin B_{t^{\prime },dist(t,t^{\prime })/2}$.

So, $B_{t,dist(t,t^{\prime })/2}\cap B_{t^{\prime },dist(t,t^{\prime })/2}=\mathrm{\varnothing }$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1056: Derivation of Tensor Product of Tensors by Real Parameter Satisfies Leibniz Rule

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that derivation of tensor product of tensors by real parameter satisfies Leibniz rule

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of tensor product of tensors.
• The reader knows a definition of derivative of real-1-parameter family of vectors in finite-dimensional real vectors space.
• The reader knows a definition of convergence of map from topological space minus point into topological space with respect to point.
• The reader admits the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences.
• The reader admits the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.

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Target Context

• The reader will have a description and a proof of the proposition that the derivation of the tensor product of any tensors by any real parameter satisfies the Leibniz rule.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $=(r_1,r_2)\subseteq \mathbb{R}$, with the subspace topology with $\mathbb{R}$ as the Euclidean topological space
$r$: $\in T$
$\{V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}\}$: $\subseteq \{\text{ the finite-dimensional }\mathbb{R}\text{ vectors spaces }\}$
$L(V_{1,1},...,V_{1,k_1}:\mathbb{R})$: $=\text{ the tensors space }$, with the canonical topology
$L(V_{2,1},...,V_{2,k_2}:\mathbb{R})$: $=\text{ the tensors space }$, with the canonical topology
$t^1$: $:T\to L(V_{1,1},...,V_{1,k_1}:\mathbb{R})$
$t^2$: $:T\to L(V_{2,1},...,V_{2,k_2}:\mathbb{R})$
$L(V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}:\mathbb{R})$: $=\text{ the tensors space }$, with the canonical topology
$t^1\otimes t^2$: $:T\to L(V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}:\mathbb{R})$
//

Statements:
$\mathrm{\exists }d{t}^1/dr\wedge \mathrm{\exists }d{t}^2/dr$
$⟹$
$\mathrm{\exists }d(t^1\otimes t^2)/dr\wedge d(t^1\otimes t^2)/dr=d{t}^1/dr\otimes t^2(r)+t^1(r)\otimes d{t}^2/dr$
//

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2: Proof

Whole Strategy: Step 1: for each of $\{V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}\}$, take any basis, $B_{j,l}=\{{b_{j,l}}_{m_{j,l}}\}$, and take the standard bases for $L(V_{1,1},...,V_{1,k_1}:\mathbb{R})$, $L(V_{2,1},...,V_{2,k_2}:\mathbb{R})$, and $L(V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}:\mathbb{R})$, $B_1$, $B_2$, and $B$; Step 2: let $t^1(r^{\prime })$, $t^2(r^{\prime })$, and $t^1(r^{\prime })\otimes t^2(r^{\prime })$ be expressed with the bases; Step 3: take $d(t^1\otimes t^2)/dr=li{m}_{r^{\prime }\to r}(t^1(r^{\prime })\otimes t^2(r^{\prime })-t^1(r)\otimes t^2(r))/(r^{\prime }-r)$ and apply the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences.

Step 1:

For each of $\{V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}$, let us take any basis, $B_{j,l}=\{{b_{j,l}}_{m_{j,l}}\}$.

Let us take the standard basis for $L(V_{1,1},...,V_{1,k_1}:\mathbb{R})$, $B_1=\{b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\}$, which is possible by the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces.

Let us take the standard basis for $L(V_{2,1},...,V_{2,k_2}:\mathbb{R})$, $B_2=\{b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}\}$, which is possible likewise.

Let us take the standard basis for $L(V_{1,1},...,V_{1,k_1},V_{2,1},...,V_{2,k_2}:\mathbb{R})$, $B=\{b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}\}$, which is possible likewise.

Step 2:

$t^1(r^{\prime })=t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}$.

$t^2(r^{\prime })=t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}$.

So, $t^1(r^{\prime })\otimes t^2(r^{\prime })=(t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}})\otimes (t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}})$.

By the property of tensor product of tensors mentioned in Note for the definition of tensor product of tensors, $=t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}$, which is the expansion of $t^1(r^{\prime })\otimes t^2(r^{\prime })$ with respect to $B$.

Step 3:

$d(t^1\otimes t^2)/dr=li{m}_{r^{\prime }\to r}(t^1(r^{\prime })\otimes t^2(r^{\prime })-t^1(r)\otimes t^2(r))/(r^{\prime }-r)$.

$=li{m}_{r^{\prime }\to r}(t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}-t_{m_{1,1},...,m_{1,k_1}}^1(r)t_{m_{2,1},...,m_{2,k_2}}^2(r)b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}})/(r^{\prime }-r)=li{m}_{r^{\prime }\to r}(t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })-t_{m_{1,1},...,m_{1,k_1}}^1(r)t_{m_{2,1},...,m_{2,k_2}}^2(r))/(r^{\prime }-r)b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}$.

By the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences, $d{t}_{m_{1,1},...,m_{1,k_1}}^1/dr=li{m}_{r^{\prime }\to r}(t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })-t_{m_{1,1},...,m_{1,k_1}}^1(r))/(r^{\prime }-r)$ and $d{t}_{m_{2,1},...,m_{2,k_2}}^2/dr=li{m}_{r^{\prime }\to r}(t_{m_{2,1},...,m_{2,k_2}}^2(r^{\prime })-t_{m_{2,1},...,m_{2,k_2}}^2(r))/(r^{\prime }-r)$ exist.

By the well-known fact in real analysis, $d(t_{m_{1,1},...,m_{1,k_1}}^1(r)t_{m_{2,1},...,m_{2,k_2}}^2(r))/dr=li{m}_{r^{\prime }\to r}(t_{m_{1,1},...,m_{1,k_1}}^1(r^{\prime })t_{m_{2,2},...,m_{2,k_2}}^2(r^{\prime })-t_{m_{1,1},...,m_{1,k_1}}^1(r)t_{m_{2,1},...,m_{2,k_2}}^2(r))/(r^{\prime }-r)$ exists and equals $d{t}_{m_{1,1},...,m_{1,k_1}}^1/dr{t}_{m_{2,1},...,m_{2,k_2}}^2(r)+t_{m_{1,1},...,m_{1,k_1}}^{1}d{t}_{m_{2,1},...,m_{2,k_2}}^2/dr$.

By the proposition that for any map from any topological space minus any point into any finite-dimensional real vectors space with the canonical topology, the convergence of the map with respect to the point exists if and only if the convergences of the component maps (with respect to any basis) with respect to the point exist, and then, the convergence Is expressed with the convergences, $d(t^1\otimes t^2)/dr=li{m}_{r^{\prime }\to r}(t^1(r^{\prime })\otimes t^2(r^{\prime })-t^1(r)\otimes t^2(r))/(r^{\prime }-r)$ exists and equals $(d{t}_{m_{1,1},...,m_{1,k_1}}^1/dr{t}_{m_{2,1},...,m_{2,k_2}}^2(r)+t_{m_{1,1},...,m_{1,k_1}}^1(r)d{t}_{m_{2,1},...,m_{2,k_2}}^2/dr)b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}$.

$=d{t}_{m_{1,1},...,m_{1,k_1}}^1/dr{t}_{m_{2,1},...,m_{2,k_2}}^2(r)b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}+t_{m_{1,1},...,m_{1,k_1}}^1(r)d{t}_{m_{2,1},...,m_{2,k_2}}^2/dr{b}_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}}\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}}=(d{t}_{m_{1,1},...,m_{1,k_1}}^1/dr{b}_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}})\otimes (t_{m_{2,1},...,m_{2,k_2}}^2(r)b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}})+(t_{m_{1,1},...,m_{1,k_1}}^1(r)b_{1,1}^{m_{1,1}}\otimes ...\otimes b_{1,k_1}^{m_{1,k_1}})\otimes (d{t}_{m_{2,1},...,m_{2,k_2}}^2/dr\otimes b_{2,1}^{m_{2,1}}\otimes ...\otimes b_{2,k_2}^{m_{2,k_2}})=d{t}^1/dr\otimes t^2(r)+t^1(r)\otimes d{t}^2/dr$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1055: For Finite-Product C∞ Manifold with Boundary and 'Vectors Spaces - Linear Morphisms' Isomorphism from Tangent Vectors Space onto Direct Sum of Tangent Vectors Spaces, Tangent Vector Operates on Function as Sum of Vectors on Projected Functions

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for finite-product $C^{\mathrm{\infty }}$ manifold with boundary and 'vectors spaces - linear morphisms' isomorphism from tangent vectors space onto direct sum of tangent vectors spaces, tangent vector operates on function as sum of vectors on projected functions

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ map projected from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing domain components except $j$-th based on point.
• The reader admits the proposition that for any finite-product $C^{\mathrm{\infty }}$ manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-product $C^{\mathrm{\infty }}$ manifold with boundary and the 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at each point onto the direct sum of the corresponding tangent vectors spaces of the constituents, any tangent vector operates on any function as the sum of the corresponding vectors on the projected functions.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$\{M_1,...,M_{n-1}\}$: $\subseteq \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_n$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_1\times ...\times M_n$: $=\text{ the finite-product }{C}^{\mathrm{\infty }}\text{ manifold with boundary }$
$m$: $=(m^1,...,m^n)\in M_1\times ...\times M_n$
$g$: $:T_m(M_1\times ...\times M_n)\to T_{m^1}{M}_1\oplus ...\oplus T_{m^n}{M}_n,v_m\mapsto (d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$, $=\text{ the canonical 'vectors spaces - linear morphisms' isomorphism }$
$v_m$: $\in T_m(M_1\times ...\times M_n)$
$f$: $\in C^{\mathrm{\infty }}(M_1\times ...\times M_n)$
//

Statements:
$v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$
//

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2: Note

We know that $v_m$ corresponds to $(d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$, but so what? I mean, how can we get $v_{m}f$ with respect to $(d{\pi }_1{v}_m,...,d{\pi }_n{v}_m)$? Certainly, $d{\pi }_j{v}_m$ cannot operate on $f$, because $f$ is not any function on $M_j$. So, what?, which is the motivation of this proposition.

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3: Proof

Whole Strategy: Step 1: take any chart around $m$, $(U_m:=U_{m^1}\times ...\times U_{m^n}\subseteq M_1\times ...\times M_n,{\varphi }_m:={\varphi }_{m^1}\times ...\times {\varphi }_{m^n})$, and let $v_m=v_m^{1,j_1}\partial /\partial x^{1,j_1}+...+v_m^{n,j_n}\partial /\partial x^{n,j_n}$; Step 2: see that $v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$.

Step 1:

Let us take any chart around $m$, $(U_m:=U_{m^1}\times ...\times U_{m^n}\subseteq M_1\times ...\times M_n,{\varphi }_m:={\varphi }_{m^1}\times ...\times {\varphi }_{m^n})$, where $(U_{m^j}\subseteq M_j,{\varphi }_{m^j})$ is a chart around $m^j$ for $M_j$, which is possible by the definition of finite-product $C^{\mathrm{\infty }}$ manifold with boundary.

$v_m=v_m^{1,j_1}\partial /\partial x^{1,j_1}+...+v_m^{n,j_n}\partial /\partial x^{n,j_n}$.

Step 2:

$v_{m}f=v_m^{1,j_1}\partial f/\partial x^{1,j_1}+...+v_m^{n,j_n}\partial f/\partial x^{n,j_n}$.

$(d{\pi }_l{v}_m)f_{l,m}=v_m(f_{l,m}\circ {\pi }_l)=v_m^{1,j_1}\partial (f_{l,m}\circ {\pi }_l)/\partial x^{1,j_1}+...+v_m^{n,j_n}(\partial f_{l,m}\circ {\pi }_l)/\partial x^{n,j_n}$.

$v_m^{o,j_o}\partial (f_{l,m}\circ {\pi }_l)/\partial x^{o,j_o}=v_m^{o,j_o}{\partial }_{o,j_o}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})$, where ${\partial }_{o,j_o}$ is the partial derivative by the $o,j_o$ component.

$f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1}$ is $(x^{\prime 1,1},...,x^{\prime 1,d_1},...,x^{\prime l,1},...,x^{\prime l,d_l},...,x^{\prime n,1},...,x^{\prime n,d_n})\mapsto (m^{\prime 1},...,m^{\prime l},...,m^{\prime n})\mapsto m^{\prime l}\mapsto f_{l,m}(m^{\prime l})=f(m^1,...,m^{\prime l},...,m^n)$.

So, when $o\ne l$, $v_m^{o,j_o}{\partial }_{o,j_o}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})=0$, because $f(m^1,...,m^{\prime l},...,m^n)$ does not depend on $(x^{\prime o,1},...,x^{\prime o,d_o})$.

So, $(d{\pi }_l{v}_m)f_{l,m}=v_m^{l,j_l}{\partial }_{l,j_l}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1})$.

On the other hand, $v_m^{l,j_l}\partial f/\partial x^{l,j_l}=v_m^{l,j_l}{\partial }_{l,j_l}(f\circ {{\varphi }_m}^{-1})$.

$f\circ {{\varphi }_m}^{-1}$ is $(x^{\prime 1,1},...,x^{\prime 1,d_1},...,x^{\prime l,1},...,x^{\prime l,d_l},...,x^{\prime n,1},...,x^{\prime n,d_n})\mapsto (m^{\prime 1},...,m^{\prime l},...,m^{\prime n})\mapsto f(m^{\prime 1},...,m^{\prime l},...,m^{\prime n})$.

While the difference between $f(m^1,...,m^{\prime l},...,m^n)$ and $f(m^{\prime 1},...,m^{\prime l},...,m^{\prime n})$ is whether $(m^{\prime 1},...,\widehat{m^{\prime l}},...,m^{\prime n})$ are fixed, $v_m^{l,j_l}{\partial }_{l,j_l}(f\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$ equals $v_m^{l,j_l}{\partial }_{l,j_l}(f_{l,m}\circ {\pi }_l\circ {{\varphi }_m}^{-1}){|}_{{\varphi }_m(m)}$, because ${\partial }_{l,j_l}$ moves only $m^{\prime l}$ anyway.

So, $(d{\pi }_l{v}_m)f_{l,m}=v_m^{l,j_l}\partial f/\partial x^{l,j_l}$.

So, $v_{m}f=\sum _{j\in \{1,...,n\}}(d{\pi }_j{v}_m)f_{j,m}$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1054: C∞ Map Projected from C∞ Map from Finite-Product C∞ Manifold with Boundary by Fixing Domain Components Except j-th Based on Point

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of $C^{\mathrm{\infty }}$ map projected from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing domain components except $j$-th based on point

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Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

---

Starting Context

• The reader knows a definition of finite-product $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

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Target Context

• The reader will have a definition of $C^{\mathrm{\infty }}$ map projected from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing domain components except $j$-th based on point.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$\{M_1,...,M_{n-1}\}$: $\subseteq \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_n$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_1\times ...\times M_n$: $=\text{ the finite-product }{C}^{\mathrm{\infty }}\text{ manifold with boundary }$
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\times ...\times M_n\to M$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$m_0$: $\in M_1\times ...\times M_n$
$j$: $\in \{1,...,n\}$
$\ast f_{j,m_0}$: $:M_j\to M,m\mapsto f(m_0^1,...,m,...,m_0^n)$, where $m$ is for the $j$-th slot
//

Conditions:
//

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2: Note

Let us see that $f_{j,m_0}$ is indeed $C^{\mathrm{\infty }}$.

Let $m\in M_j$ be any.

$\tilde{m}:=(m_0^1,...,m,...,m_0^n)=({\tilde{m}}^1,...,{\tilde{m}}^n)$.

As $f$ is $C^{\mathrm{\infty }}$ at $\tilde{m}$, there are a chart around $\tilde{m}$, $(U_{\tilde{m}}\subseteq M_1\times ...\times M_n,{\varphi }_{\tilde{m}})$, and a chart around $f(\tilde{m})$, $(U_{f(\tilde{m})}\subseteq M,{\varphi }_{f(\tilde{m})})$, such that $f(U_{\tilde{m}})\subseteq U_{f(\tilde{m})}$ and ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}:{\varphi }_{\tilde{m}}(U_{\tilde{m}})\to {\varphi }_{f(\tilde{m})}(U_{f(\tilde{m})})$ is $C^{\mathrm{\infty }}$ at ${\varphi }_{\tilde{m}}(\tilde{m})$.

By the of finite-product $C^{\mathrm{\infty }}$ manifold with boundary, $(U_{\tilde{m}}\subseteq M_1\times ...\times M_n,{\varphi }_{\tilde{m}})$ can be chosen as $U_{\tilde{m}}=U_{1,{\tilde{m}}^1}\times ...\times U_{n,{\tilde{m}}^n}$ and ${\varphi }_{\tilde{m}}={\varphi }_{1,{\tilde{m}}^1}\times ...\times {\varphi }_{n,{\tilde{m}}^n}$, where $(U_{j,{\tilde{m}}^j}\subseteq M_j,{\varphi }_{j,{\tilde{m}}^j})$ is a chart for $M_j$, while $U_{\tilde{m}}=U_{1,{m_0}^1}\times ...\times U_{j,m}\times ...\times U_{n,{m_0}^n}$ and ${\varphi }_{\tilde{m}}={\varphi }_{1,{m_0}^1}\times ...\times {\varphi }_{j,m}\times ...\times {\varphi }_{n,{m_0}^n}$.

Obviously, ${{\varphi }_{\tilde{m}}}^{-1}={{\varphi }_{1,{m_0}^1}}^{-1}\times ...\times {{\varphi }_{j,m}}^{-1}\times ...\times {{\varphi }_{n,{m_0}^n}}^{-1}$.

$(U_{j,m}\subseteq M_j,{\varphi }_{j,m})$ is the chart around $m$.

$f_{S,m_0}(U_{j,m})\subseteq U_{f(\tilde{m})}$, because for each $m^{\prime }\in U_{j,m}$, $f_{S,m_0}(m^{\prime })=f((m_0^1,...,m^{\prime },...,m_0^n))$, but $(m_0^1,...,m^{\prime },...,m_0^n)\in U_{1,{m_0}^1}\times ...\times U_{j,m}\times ...\times U_{n,{m_0}^n}=U_{\tilde{m}}$ while $f(U_{\tilde{m}})\subseteq U_{f(\tilde{m})}$.

Let us think of ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_{j,m}}^{-1}:{\varphi }_{j,m}(U_{j,m})\to {\varphi }_{f(\tilde{m})}(U_{f(\tilde{m})})$.

It is ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_{j,m}}^{-1}:x^{\prime }\mapsto {\varphi }_{f(\tilde{m})}\circ f_{j,m_0}(m^{\prime })={\varphi }_{f(\tilde{m})}\circ f((m_0^1,...,m^{\prime },...,m_0^n))$.

That is in fact same with ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}$ where $({\varphi }_{{\tilde{m}}^1}(m_0^1),...,\widehat{x^{\prime }},...,{\varphi }_{{\tilde{m}}^n}(m_0^n))$ is fixed.

As ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}$ is $C^{\mathrm{\infty }}$ at ${\varphi }_{\tilde{m}}(\tilde{m})$, it is $C^{\mathrm{\infty }}$ with respect to $x^{\prime }$, so, ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_{j,m}}^{-1}$ is $C^{\mathrm{\infty }}$ with respect to $x^{\prime }$.

So, $f_{S,m_0}$ is $C^{\mathrm{\infty }}$ at $m$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1053: C∞ Map Induced from C∞ Map from Finite-Product C∞ Manifold with Boundary by Fixing Some Domain Components Based on Point

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of $C^{\mathrm{\infty }}$ map induced from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing some domain components based on point

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

---

Starting Context

• The reader knows a definition of finite-product $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

---

Target Context

• The reader will have a definition of $C^{\mathrm{\infty }}$ map induced from $C^{\mathrm{\infty }}$ map from finite-product $C^{\mathrm{\infty }}$ manifold with boundary by fixing some domain components based on point.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$\{M_1,...,M_{n-1}\}$: $\subseteq \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_n$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_1\times ...\times M_n$: $=\text{ the finite-product }{C}^{\mathrm{\infty }}\text{ manifold with boundary }$
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\times ...\times M_n\to M$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$m_0$: $\in M_1\times ...\times M_n$
$S$: $=\{j_1,...,j_s\}\subseteq \{1,...,n\}$
$\ast f_{S,m_0}$: $:M_1\times ...\times M_n\to M,m=(m^1,...m^n)\mapsto f(m_0^1,...,m^{j_1},...,m^{j_s},...,m_0^n)$, which means that $\{m_0^1,...,m_0^n\}$ are used except $\{m^{j_1},...,m^{j_s}\}$
//

Conditions:
//

When $S=\{j\}$, $f_{S,m_0}$ is denoted also as $f_{j,m_0}$, which is a typical case.

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2: Note

Let us see that $f_{S,m_0}$ is indeed $C^{\mathrm{\infty }}$.

Let $m=(m^1,...,m^n)\in M_1\times ...\times M_n$ be any.

$\tilde{m}:=(m_0^1,...,m^{j_1},...,m^{j_s},...,m_0^n)=({\tilde{m}}^1,...,{\tilde{m}}^n)$.

As $f$ is $C^{\mathrm{\infty }}$ at $\tilde{m}$, there are a chart around $\tilde{m}$, $(U_{\tilde{m}}\subseteq M_1\times ...\times M_n,{\varphi }_{\tilde{m}})$, and a chart around $f(\tilde{m})$, $(U_{f(\tilde{m})}\subseteq M,{\varphi }_{f(\tilde{m})})$, such that $f(U_{\tilde{m}})\subseteq U_{f(\tilde{m})}$ and ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}:{\varphi }_{\tilde{m}}(U_{\tilde{m}})\to {\varphi }_{f(\tilde{m})}(U_{f(\tilde{m})})$ is $C^{\mathrm{\infty }}$ at ${\varphi }_{\tilde{m}}(\tilde{m})$.

By the definition of finite-product $C^{\mathrm{\infty }}$ manifold with boundary, $(U_{\tilde{m}}\subseteq M_1\times ...\times M_n,{\varphi }_{\tilde{m}})$ can be chosen as $U_{\tilde{m}}=U_{1,{\tilde{m}}^1}\times ...\times U_{n,{\tilde{m}}^n}$ and ${\varphi }_{\tilde{m}}={\varphi }_{1,{\tilde{m}}^1}\times ...\times {\varphi }_{n,{\tilde{m}}^n}$, where $(U_{j,{\tilde{m}}^j}\subseteq M_j,{\varphi }_{j,{\tilde{m}}^j})$ is a chart for $M_j$. $U_{\tilde{m}}=U_{1,{m_0}^1}\times ...\times U_{j_1,m^{j_1}}\times ...\times U_{j_s,m^{j_s}}\times ...\times U_{n,{m_0}^n}$ and ${\varphi }_{\tilde{m}}={\varphi }_{1,{m_0}^1}\times ...\times {\varphi }_{j_1,m^{j_1}}\times ...\times {\varphi }_{j_s,m^{j_s}}\times ...\times {\varphi }_{n,{m_0}^n}$.

Obviously, ${{\varphi }_{\tilde{m}}}^{-1}={{\varphi }_{1,{m_0}^1}}^{-1}\times ...\times {{\varphi }_{j_1,m^{j_1}}}^{-1}\times ...\times {{\varphi }_{j_s,m^{j_s}}}^{-1}\times ...\times {{\varphi }_{n,{m_0}^n}}^{-1}$.

Let us take any chart around $m$, $(U_m\subseteq M_1\times ...\times M_n,{\varphi }_m)$, such that $U_m:=U_{1,m^1}\times ...\times U_{j_1,m^{j_1}}\times ...\times U_{j_s,m^{j_s}}\times ...U_{n,m^n}$ and ${\varphi }_m={\varphi }_{1,m^1}\times ...\times {\varphi }_{j_1,m^{j_1}}\times ...\times {\varphi }_{j_s,m^{j_s}}\times ...\times {\varphi }_{n,m^n}$, where $(U_{j,m^j}\subseteq M_j,{\varphi }_{j,m^j})$ is a chart for $M_j$, while $U_{j_1,m^{j_1}},...,U_{j_s,m^{j_s}}$ and ${\varphi }_{j_1,m^{j_1}},...,{\varphi }_{j_s,m^{j_s}}$ are the ones introduced above.

$f_{S,m_0}(U_m)\subseteq U_{f(\tilde{m})}$, because for each $m^{\prime }=(m^{\prime 1},...,m^{\prime n})\in U_m$, $f_{S,m_0}(m^{\prime })=f((m_0^1,...,m^{\prime j_1},...,m^{\prime j_s},...,m_0^n))$, but $(m_0^1,...,m^{\prime j_1},...,m^{\prime j_s},...,m_0^n)\in U_{1,{m_0}^1}\times ...\times U_{j_1,m^{j_1}}\times ...\times U_{j_s,m^{j_s}}\times ...\times U_{n,{m_0}^n}=U_{\tilde{m}}$ while $f(U_{\tilde{m}})\subseteq U_{f(\tilde{m})}$.

Let us think of ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_m}^{-1}:{\varphi }_m(U_m)\to {\varphi }_{f(\tilde{m})}(U_{f(\tilde{m})})$.

It is ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_{1,m^1}}^{-1}\times ...\times {{\varphi }_{j_1,m^{j_1}}}^{-1}\times ...\times {{\varphi }_{j_s,m^{j_s}}}^{-1}\times ...\times {{\varphi }_{n,m^n}}^{-1}:(x^{\prime 1},...,x^{\prime n})\mapsto {\varphi }_{f(\tilde{m})}\circ f_{S,m_0}((m^{\prime 1},...,m^{\prime n}))={\varphi }_{f(\tilde{m})}\circ f((m_0^1,...,m^{\prime j_1},...,m^{\prime j_s},...,m_0^n))$.

That is in fact same with ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}$ where $({\varphi }_{{\tilde{m}}^1}(m_0^1),...,\widehat{x^{\prime j_1}},...,\widehat{x^{\prime j_s}},...,{\varphi }_{{\tilde{m}}^n}(m_0^n))$ is fixed.

As ${\varphi }_{f(\tilde{m})}\circ f\circ {{\varphi }_{\tilde{m}}}^{-1}$ is $C^{\mathrm{\infty }}$ at ${\varphi }_{\tilde{m}}(\tilde{m})$, it is $C^{\mathrm{\infty }}$ with respect to $(x^{\prime j_1},...,x^{\prime j_s})$, so, ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_m}^{-1}$ is $C^{\mathrm{\infty }}$ with respect to $(x^{\prime j_1},...,x^{\prime j_s})$.

${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_m}^{-1}$ is constant with respect to $(x^{\prime 1},...,\widehat{x^{\prime j_1}},...,\widehat{x^{\prime j_s}},...,x^{\prime n})$, so, is $C^{\mathrm{\infty }}$ with respect to $(x^{\prime 1},...,\widehat{x^{\prime j_1}},...,\widehat{x^{\prime j_s}},...,x^{\prime n})$.

So, ${\varphi }_{f(\tilde{m})}\circ f_{S,m_0}\circ {{\varphi }_m}^{-1}$ is $C^{\mathrm{\infty }}$ with respect $x^{\prime 1},...,x^{\prime n}$.

So, $f_{S,m_0}$ is $C^{\mathrm{\infty }}$ at $m$.

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References

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1052: For Continuous Map Between Topological Spaces and Subset of Domain Mapped into Open Subset of Codomain, There Is Open Neighborhood of Domain Subset Mapped into Open Subset

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description/proof of that for continuous map between topological spaces and subset of domain mapped into open subset of codomain, there is open neighborhood of domain subset mapped into open subset

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces and any subset of the domain that is mapped into any open subset of the codomain, there is an open neighborhood of the domain subset mapped into the codomain open subset.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S_1$: $\subseteq T_1$
$U_2$: $\in \{\text{ the open subsets of }{T}_2\}$
//

Statements:
$f(S_1)\subseteq U_2$
$⟹$
$\mathrm{\exists }{U}_1\subseteq T_1\in \{\text{ the open subsets of }{T}_1\}(S_1\subseteq U_1\wedge f(U_1)\subseteq U_2)$
//

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2: Proof

Whole Strategy: Step 1: for each $s\in S_1$, take an open neighborhood of $s$, $U_s\subseteq T_1$, such that $f(U_s)\subseteq U_2$; Step 2: take $U_1:={\cup }_{s\in S_1}{U}_s$ and see that $f(U_1)\subseteq U_2$.

Step 1:

Let $s\in S_1$ be any.

$f(s)\in U_2$, which means that $U_2$ is an open neighborhood of $f(s)$.

As $f$ is continuous, there is an open neighborhood of $s$, $U_s\subseteq T_1$, such that $f(U_s)\subseteq U_2$.

Step 2:

Let us define $U_1:={\cup }_{s\in S_1}{U}_s$, which is open on $T_1$.

$S_1\subseteq U_1$, because for each $s\in S_1$, $s\in U_s\subseteq U_1$.

For each $u\in U_1$, $u\in U_s$ for an $s$. So, $f(u)\in U_2$ because $f(U_s)\subseteq U_2$.

So, $f(U_1)\subseteq U_2$.

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References

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