1182: For Topological Space and Subset of Subspace, Closure of Subset on Subspace Is Contained in Closure of Subset on Base Space

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

description/proof of that for topological space and subset of subspace, closure of subset on subspace is contained in closure of subset on base space

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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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is contained in the closure of the subset on the base space.

---

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:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\in \{\text{ the topological subspaces of }{T}^{\prime }\}$
$S$: $\subseteq T$
//

Statements:
${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$, where ${\bar{S}}^T$ is the closure of $S$ on $T$ and ${\bar{S}}^{T^{\prime }}$ is the closure of $S$ on $T^{\prime }$
//

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2: Note

The equality does not necessarily hold.

For example, let $T^{\prime }=\mathbb{R}$ with the Euclidean topology, $T=(-1,1)$, and $S=(-1,1)$, then, ${\bar{S}}^T=(-1,1)\subset [-1,1]={\bar{S}}^{T^{\prime }}$.

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3: Proof

Whole Strategy: Step 1: see that $S\subseteq {\bar{S}}^{T^{\prime }}\cap T$ and ${\bar{S}}^{T^{\prime }}\cap T$ is closed on $T$, and see that ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T\subseteq {\bar{S}}^{T^{\prime }}$.

Step 1:

$S\subseteq {\bar{S}}^{T^{\prime }}$.

As $S\subseteq T$, $S\subseteq {\bar{S}}^{T^{\prime }}\cap T$.

As ${\bar{S}}^{T^{\prime }}$ is closed on $T^{\prime }$, ${\bar{S}}^{T^{\prime }}\cap T$ is closed on $T$, by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.

As ${\bar{S}}^T$ is the intersection of all the closed subsets of $T$ that contain $S$, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T$.

But ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T\subseteq {\bar{S}}^{T^{\prime }}$.

---

References

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

1181: For Continuous Map Between Topological Spaces, Image of Closure of Subset Is Contained in Closure of Image of Subset

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

description/proof of that for continuous map between topological spaces, image of closure of subset is contained in closure of image of subset

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of continuous map.
• The reader knows a definition of closure of subset of topological space.
• 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.

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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, the image of the closure of any subset of the domain is contained in the closure of the image of the subset.

---

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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S_1$: $\subseteq T_1$
//

Statements:
$f(\overline{S_1})\subseteq \overline{f(S_1)}$
//

---

2: Proof

Whole Strategy: Step 1: let $p\in \overline{S_1}$ be any and see that $p\in S_1$ or $p\in \overline{S_1}\setminus S_1$; Step 2: when $p\in S_1$, see that $f(p)\in f(S_1)\subseteq \overline{f(S_1)}$; Step 3: when $p\in \overline{S_1}\setminus S_1$, see that $f(p)\in \overline{f(S_1)}$.

Step 1:

Let $p\in \overline{S_1}$ be any.

$p\in S_1$ or $p\in \overline{S_1}\setminus S_1$.

Step 2:

Let us suppose that $p\in S_1$.

$f(p)\in f(S_1)\subseteq \overline{f(S_1)}$.

Step 3:

Let us suppose that $p\in \overline{S_1}\setminus S_1$.

That means that $p$ is an accumulation point of $S_1$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

$f(p)\in f(S_1)$ (which is possible because $f$ is not supposed to be injective) or $f(p)\notin f(S_1)$.

When $f(p)\in f(S_1)$, $f(p)\in f(S_1)\subseteq \overline{f(S_1)}$.

Let us suppose that $f(p)\notin f(S_1)$ hereafter.

Let any open neighborhood of $f(p)$ be $U_{f(p)}\subseteq T_2$.

As $f$ is continuous, $f^{-1}(U_{f(p)})\subseteq T_1$ is open, and contains $p$, so, is an open neighborhood of $p$.

As $p$ is an accumulation point of $S_1$, there is a point, $p^{\prime }\in f^{-1}(U_{f(p)})\cap S_1$.

$f(p^{\prime })\in U_{f(p)}\cap f(S_1)$, which implies that $f(p)$ is an accumulation point of $f(S_1)$.

So, $f(p)\in \overline{f(S_1)}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, $f(p)\in \overline{f(S_1)}$ anyway.

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3: Note

If $f$ is a homeomorphism, $f(\overline{S_1})=\overline{f(S_1)}$, because $f^{-1}$ is a continuous map and $f^{-1}(f(S_1))=S_1$, and so, $f^{-1}(\overline{f(S_1)})\subseteq \overline{f^{-1}(f(S_1))}=\overline{S_1}$, and so, $f(f^{-1}(\overline{f(S_1)}))=\overline{f(S_1)}\subseteq f(\overline{S_1})$.

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References

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

1180: For Net with Directed Index Set on Subset of Topological Space That Converges on Space, Convergence Is on Closure of Subset

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

description/proof of that for net with directed index set on subset of topological space that converges on space, convergence is on closure of subset

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of convergence of net with directed index set.
• The reader knows a definition of closure of subset of topological space.
• The reader admits a local characterization of closure: any point on any topological space is on the closure of any subset if and only if its every neighborhood intersects the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any net with directed index set on any subset of any topological space that converges on the space, the convergence is on the closure of the subset.

---

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:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$D$: $\in \{\text{ the directed index sets }\}$
$N$: $:D\to T$, $N(D)\subseteq S$
$p$: $\in T$
//

Statements:
$N$ converges to $p$
$⟹$
$p\in \bar{S}$.
//

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2: Proof

Whole Strategy: Step 1: suppose that $p\notin \bar{S}$ and find a contradiction.

Step 1:

Let us suppose that $N$ converges to $p$.

Let us suppose that $p\notin \bar{S}$.

By a local characterization of closure: any point on any topological space is on the closure of any subset if and only if its every neighborhood intersects the subset, there would be a neighborhood of $p$, $N_p\subseteq T$, such that $N_p\cap S=\mathrm{\varnothing }$.

But there would be an index, $j_0\in D$, such that $N(j)\in N_p$ for every $j\in D$ such that $j_0\le j$, but $N(j)\in S$, which would mean that $N_p\cap S\ne \mathrm{\varnothing }$, a contradiction.

So, $p\in \bar{S}$.

---

References

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

1179: For Disjoint Union Topological Space, Closure of Disjoint Union of Subsets Is Disjoint Union of Closures of Subsets

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

description/proof of that for disjoint union topological space, closure of disjoint union of subsets is disjoint union of closures of subsets

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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

---

Starting Context

• The reader knows a definition of disjoint union topology.
• The reader knows a definition of closure of subset of topological space.
• 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.

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Target Context

• The reader will have a description and a proof of the proposition that for any disjoint union topological space, the closure of the disjoint union of any subsets is the disjoint union of the closures of the subsets.

---

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:
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{T_j|j\in J\}$: $T_j\in \{\text{ the topological spaces }\}$
$T$: $=\coprod _{j\in J}{T}_j$
$\{S_j\subseteq T_j|j\in J\}$:
//

Statements:
$\overline{\coprod _{j\in J}{S}_j}=\coprod _{j\in J}\overline{S_j}$, where the 1st overline is the closure on $T$ and the 2nd overline is the closure on $T_j$
//

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2: Proof

Whole Strategy: Step 1: see that $\overline{\coprod _{j\in J}{S}_j}\subseteq \coprod _{j\in J}\overline{S_j}$; Step 2: see that $\coprod _{j\in J}\overline{S_j}\subseteq \overline{\coprod _{j\in J}{S}_j}$.

Step 1:

$\coprod _{j\in J}{S}_j\subseteq \coprod _{j\in J}\overline{S_j}$, and the right hand side is closed on $T$, by the definition of disjoint union topology.

So, $\overline{\coprod _{j\in J}{S}_j}\subseteq \coprod _{j\in J}\overline{S_j}$, by the definition of closure of subset of topological space.

Step 2:

Let $p\in \coprod _{j\in J}\overline{S_j}$ be any.

$p\in \overline{S_j}$ for a $j$.

Let $U_p\subseteq T$ be any open neighborhood of $p$.

$U_p\cap T_j$ is open on $T_j$, by the definition of disjoint union topology, and $p\in U_p\cap T_j$.

So, $U_p\cap T_j$ is an open neighborhood of $p$ on $T_j$.

$U_p\cap T_j\cap S_j\ne \mathrm{\varnothing }$, because $p\in \overline{S_j}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, $U_p\cap \coprod _{j\in J}{S}_j\ne \mathrm{\varnothing }$.

So, $p\in \overline{\coprod _{j\in J}{S}_j}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, $\coprod _{j\in J}\overline{S_j}\subseteq \overline{\coprod _{j\in J}{S}_j}$.

So, $\overline{\coprod _{\alpha \in A}{S}_{\alpha }}=\coprod _{\alpha \in A}\overline{S_{\alpha }}$.

---

References

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

1178: For Topological Space and Subset of Subspace, if Subspace Is Closed, Closure of Subset on Subspace Is Closure on Base Space

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

description/proof of that for topological space and subset of subspace, if subspace is closed, closure of subset on subspace is closure on base space

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader knows a definition of subspace topology of subset of topological space.
• The reader knows a definition of closed subset of topological space.
• The reader admits the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace.
• The reader admits the proposition that for any topological space and any subset of any subspace, if the closure of the subset on the subspace is closed on the base space, the closure is the closure of the subset on the base space.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, if the subspace is closed on the base space, the closure of the subset on the subspace is the closure of the subset on the base space.

---

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:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\in \{\text{ the topological subspaces of }{T}^{\prime }\}$
$S$: $\subseteq T$
//

Statements:
$T\in \{\text{ the closed subspaces of }{T}^{\prime }\}$
$⟹$
${\bar{S}}^T={\bar{S}}^{T^{\prime }}$, where ${\bar{S}}^T$ is the closure of $S$ on $T$ and ${\bar{S}}^{T^{\prime }}$ is the closure of $S$ on $T^{\prime }$
//

---

2: Proof

Whole Strategy: Step 1: see that ${\bar{S}}^T={\bar{S}}^{T^{\prime }}\cap T$; Step 2: see that ${\bar{S}}^T={\bar{S}}^{T^{\prime }}$.

Step 1:

${\bar{S}}^T={\bar{S}}^{T^{\prime }}\cap T$, by the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace.

${\bar{S}}^{T^{\prime }}\cap T$ is closed on $T^{\prime }$ as an intersection of closed subsets.

Step 2:

${\bar{S}}^T={\bar{S}}^{T^{\prime }}$, by the proposition that for any topological space and any subset of any subspace, if the closure of the subset on the subspace is closed on the base space, the closure is the closure of the subset on the base space.

---

References

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

1177: For Topological Space and Subset of Subspace, Closure of Subset on Subspace Is Intersection of Closure of Subset on Base Space and Subspace

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

description/proof of that for topological space and subset of subspace, closure of subset on subspace is intersection of closure of subset on base space and subspace

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader knows a definition of subspace topology of subset of topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace.

---

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:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\in \{\text{ the topological subspaces of }{T}^{\prime }\}$
$S$: $\subseteq T$
//

Statements:
${\bar{S}}^T={\bar{S}}^{T^{\prime }}\cap T$, where ${\bar{S}}^T$ is the closure of $S$ on $T$ and ${\bar{S}}^{T^{\prime }}$ is the closure of $S$ on $T^{\prime }$
//

---

2: Proof

Whole Strategy: Step 1: see that ${\bar{S}}^{T^{\prime }}={\cap }_{j\in J}{C}_j^{\prime }$ where $\{C_j^{\prime }|j\in J\}$ is the set of the closed subsets of $T^{\prime }$ that contain $S$; Step 2: see that ${\bar{S}}^{T^{\prime }}\cap T={\cap }_{j\in J}{C}_j^{\prime }\cap T$ is the intersection of the closed subsets of $T$ that contain $S$.

Step 1:

${\bar{S}}^{T^{\prime }}$ is the intersection of the closed subsets of $T^{\prime }$ that contain $S$.

So, letting $\{C_j^{\prime }|j\in J\}$ be the set of the closed subsets of $T^{\prime }$ that contain $S$ where $J$ is a possibly uncountable index set, ${\bar{S}}^{T^{\prime }}={\cap }_{j\in J}{C}_j^{\prime }$.

Step 2:

${\bar{S}}^{T^{\prime }}\cap T=({\cap }_{j\in J}{C}_j^{\prime })\cap T={\cap }_{j\in J}(C_j^{\prime }\cap T)$.

$C_j^{\prime }\cap T$ is a closed subset of $T$ by the definition of subspace topology and $S\subseteq C_j^{\prime }\cap T$, and any closed subset of $T$ that contains $S$, $C$, appears as $C=C_j^{\prime }\cap T$ for a $j$, because there is a closed subset of $T^{\prime }$, $C^{\prime }\subseteq T^{\prime }$, such that $C=C^{\prime }\cap T$, but $S\subseteq C\subseteq C^{\prime }$, so, $C^{\prime }$ is a closed subset of $T^{\prime }$ that contains $S$, so, $C^{\prime }=C_j^{\prime }$ for a $j$.

So, ${\cap }_{j\in J}(C_j^{\prime }\cap T)$ is the intersection of all the closed subsets of $T$ that contain $S$, which is ${\bar{S}}^T$.

So, ${\bar{S}}^T={\bar{S}}^{T^{\prime }}\cap T$.

---

References

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

1176: For Topological Space and Subset of Subspace, if Its Closure on Subspace Is Closed on Base Space, the Closure Is Closure on Base Space

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

description/proof of that for topological space and subset of subspace, if its closure on subspace is closed on base space, the closure is closure on base space

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader knows a definition of subspace topology of subset of topological space.
• The reader knows a definition of closed subset of topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, if the closure of the subset on the subspace is closed on the base space, the closure is the closure of the subset on the base space.

---

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:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\in \{\text{ the topological subspaces of }{T}^{\prime }\}$
$S$: $\subseteq T$
//

Statements:
${\bar{S}}^T\in \{\text{ the closed subsets of }{T}^{\prime }\}$, where ${\bar{S}}^T$ denotes the closure of $S$ on $T$
$⟹$
${\bar{S}}^T={\bar{S}}^{T^{\prime }}$, where ${\bar{S}}^{T^{\prime }}$ denotes the closure of $S$ on $T^{\prime }$
//

---

2: Proof

Whole Strategy: Step 1: see that ${\bar{S}}^{T^{\prime }}\subseteq {\bar{S}}^T$; Step 2: see that ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$; Step 3: conclude the proposition.

Step 1:

${\bar{S}}^{T^{\prime }}\subseteq {\bar{S}}^T$, because ${\bar{S}}^{T^{\prime }}$ is the smallest closed subset of $T^{\prime }$ that (the closed subset) contains $S$ while ${\bar{S}}^T$ is a closed subset of $T^{\prime }$ that (the closed subset) contains $S$, by the supposition.

Step 2:

$S\subseteq {\bar{S}}^{T^{\prime }}\cap T$, which is closed on $T$, so, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T$, because ${\bar{S}}^T$ is the smallest closed subset of $T$ that (the closed subset) contains $S$ while ${\bar{S}}^{T^{\prime }}\cap T$ is a closed subset of $T$ that (the closed subset) contains $S$, so, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$.

Step 3:

So, ${\bar{S}}^T={\bar{S}}^{T^{\prime }}$.

---

References

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

1175: For Topological Space, if Intersection of Subset and Open Subset Is Closed on Open Subset Subspace, Intersection Equals Intersection of Closure of Subset and Open Subset

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

description/proof of that for topological space, if intersection of subset and open subset is closed on open subset subspace, intersection equals intersection of closure of subset and open subset

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of subspace topology of subset of topological space.
• The reader knows a definition of closed subset of topological space.
• The reader knows a definition of closure of subset of topological space.
• The reader admits the proposition that for any topological space, the intersection of the closure of any subset and any open subset is contained in the closure of the intersection of the subset and the open subset.

---

Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any subset, and any open subset, if the intersection of the subset and the open subset is closed on the open subset subspace, the intersection equals the intersection of the closure of the subset and the open subset.

---

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:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$U$: $\in \{\text{ the open subsets of }T\}$ with the subspace topology
//

Statements:
$S\cap U\in \{\text{ the closed subsets of }U\}$
$⟹$
$S\cap U=\bar{S}\cap U$, where $\bar{S}$ is the closure of $S$ on $T$
//

---

2: Proof

Whole Strategy: Step 1: see that $S\cap U=\overline{S\cap U}\cap U$; Step 2: see that $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}$; Step 3: see that $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}\cap U$.

Step 1:

There is a closed subset, $C\subset T$, such that $S\cap U=C\cap U$, by the definition of subspace topology.

$S\cap U\subseteq C$ and $\overline{S\cap U}\subseteq C$ where the closure is on $T$, because the closure is the smallest closed subset that contains the concerned subset.

Then, $C$ can be taken to be $\overline{S\cap U}$, which is a closed subset of $T$, contains $S\cap U$, so, the replacement does not lose any point on $S\cap U$, and is contained in $C$, so the replacement does not add any extra point. So, $S\cap U=\overline{S\cap U}\cap U$.

Step 2:

On the other hand, $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}$, by the proposition that for any topological space, the intersection of the closure of any subset and any open subset is contained in the closure of the intersection of the subset and the open subset.

Step 3:

By taking the intersection of it with $U$, $S\cap U\cap U=S\cap U\subseteq \bar{S}\cap U\cap U=\bar{S}\cap U\subseteq \overline{S\cap U}\cap U$. But as the 1st term and the last term equal, the middle term equals them, so, $S\cap U=\bar{S}\cap U$.

---

3: Note

$U$ has to be open for this proposition: for example, if $U$ is not open, here is a counterexample: $T=\mathbb{R}$ with the Euclidean topology, $U=[0,1]$, and $S=(-1,0)$, then $S\cap U=\mathrm{\varnothing }$, closed, but $\bar{S}\cap U=[-1,0]\cap [0,1]=\{0\}$. Such a counterexample does not work for when $U$ is open, because if $U=(0,1)$, $S\cap U=(-1,0)\cap (0,1)=\mathrm{\varnothing }$ and $\bar{S}\cap U=[-1,0]\cap (0,1)=\mathrm{\varnothing }$, and if $U=(-p,1)$ for any $0<p<1$, $S\cap U=(-1,0)\cap (-p,1)=(-p,0)$, which is not closed.

---

References

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

1174: For Topological Space, Closure of Intersection of Subsets Is Contained in Intersection of Closures of Subsets

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

description/proof of that for topological space, closure of intersection of subsets is contained in intersection of closures of subsets

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any topological space, the closure of the intersection of any subsets is contained in the intersection of the closures of the subsets.

---

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:
$T$: $\in \{\text{ the topological spaces }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{S_j\subseteq T|j\in J\}$:
//

Statements:
$\overline{{\cap }_{j\in J}{S}_j}\subseteq {\cap }_{j\in J}\overline{S_j}$
//

---

2: Note

Equality does not necessarily hold even if $J$ is finite.

For example, let $T=\mathbb{R}$ with the Euclidean topology and $\{S_1=(-1,0),S_2=(0,1)\}$, then, $\overline{{\cap }_{j\in J}{S}_j}=\overline{(-1,0)\cap (0,1)}=\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }\subset {\cap }_{j\in J}\overline{S_j}=\overline{(-1,0)}\cap \overline{(0,1)}=[-1,0]\cap [0,1]=\{0\}$.

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3: Proof

Whole Strategy: Step 1: see that ${\cap }_{j\in J}{S}_j\subseteq {\cap }_{j\in J}\overline{S_j}$ and ${\cap }_{j\in J}\overline{S_j}$ is closed.

Step 1:

As $S_j\subseteq \overline{S_j}$, ${\cap }_{j\in J}{S}_j\subseteq {\cap }_{j\in J}\overline{S_j}$.

${\cap }_{j\in J}\overline{S_j}$ is closed as an intersection of closed subsets.

So, ${\cap }_{j\in J}\overline{S_j}$ is a closed subset that contains ${\cap }_{j\in J}{S}_j$.

As $\overline{{\cap }_{j\in J}{S}_j}$ is the intersection of all the closed subsets that contains ${\cap }_{j\in J}{S}_j$, $\overline{{\cap }_{j\in J}{S}_j}\subseteq {\cap }_{j\in J}\overline{S_j}$.

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References

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

1173: C∞ 1-Form Operated on C∞ Vectors Field Along C∞ Curve Is C∞ Function over Domain of Curve

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description/proof of that $C^{\mathrm{\infty }}$ 1-Form operated on $C^{\mathrm{\infty }}$ vector field along $C^{\mathrm{\infty }}$ curve is $C^{\mathrm{\infty }}$ function over domain of curve

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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: Proof

---

Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field along $C^{\mathrm{\infty }}$ curve.
• The reader knows a definition of $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that any $(0,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and any $C^{\mathrm{\infty }}$ curve on it, any $C^{\mathrm{\infty }}$ 1-form operated on any $C^{\mathrm{\infty }}$ vector field along the curve is a $C^{\mathrm{\infty }}$ function over the domain of the curve.

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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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$I$: $=(s_1,s_2)\subseteq \mathbb{R}$ as the open submanifold of the Euclidean $C^{\mathrm{\infty }}$ manifold
$\lambda$: $:I\to M$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ curves }\}$
$V$: $:I\to \lambda (I)\to TM$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors fields along }\lambda \}$
$t$: $:M\to T_1^0(TM)$, $\in \{\text{ the }{C}^{\mathrm{\infty }}1\text{ -forms }\}$
//

Statements:
$t(V):I\to \mathbb{R}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

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2: Proof

For each $s\in I$, let us take a chart, $(U_{\lambda (s)}\subseteq M,{\varphi }_{\lambda (s)})$, and the induced chart, $({\pi }^{-1}(U_{\lambda (s)})\subseteq TM,\widetilde{{\varphi }_{\lambda (s)}})$.

As $\lambda$ is continuous, there is an open neighborhood of $s$, $U_s\subseteq I$, such that $\lambda (U_s)\subseteq U_{\lambda (s)}$.

Over $U_s$, $V(\lambda (s))=V^j(\lambda (s))\partial /\partial x^j$, where $V^j(\lambda (s))$ s are $C^{\mathrm{\infty }}$ as some functions of $s$, because $V$ is $C^{\mathrm{\infty }}$.

$t(V(\lambda (s)))=t(V^j(\lambda (s))\partial /\partial x^j)=V^j(\lambda (s))t(\partial /\partial x^j)$, but as $\partial /\partial x^j$ is a $C^{\mathrm{\infty }}$ vectors field over $U_{\lambda (s)}$, $t(\partial /\partial x^j)$ is a $C^{\mathrm{\infty }}$ function over $U_{\lambda (s)}$, by the proposition that any $(0,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$, and $V^j(\lambda (s))t(\partial /\partial x^j)=V^j(\lambda (s))(t(\partial /\partial x^j))(\lambda (s))$, and $(t(\partial /\partial x^j))(\lambda (s))$ is a $C^{\mathrm{\infty }}$ function of $s$ because $\lambda$ is $C^{\mathrm{\infty }}$, so, $t(V(\lambda (s)))$ is a $C^{\mathrm{\infty }}$ function of $s$.

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References

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

1172: For C∞ Manifold with Boundary, Wedge Product of C∞ Forms Is C∞ Form

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary, wedge product of $C^{\mathrm{\infty }}$ forms is $C^{\mathrm{\infty }}$ form

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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: Proof

---

Starting Context

• The reader knows a definition of $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of wedge product of multicovectors.
• The reader admits the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, the wedge product of any $C^{\mathrm{\infty }}$ forms is a $C^{\mathrm{\infty }}$ form.

---

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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f_1$: $:M\to {\mathrm{\Lambda }}_{q_1}(TM)$, $\in \{\text{ the }{C}^{\mathrm{\infty }}{q}_1\text{ -forms }\}$
$f_2$: $:M\to {\mathrm{\Lambda }}_{q_2}(TM)$, $\in \{\text{ the }{C}^{\mathrm{\infty }}{q}_2\text{ -forms }\}$
//

Statements:
$f_1\wedge f_2\in \{\text{ the }{C}^{\mathrm{\infty }}(q_1+q_2)\text{ -forms }\}$
//

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2: Proof

Whole Strategy: Step 1: apply the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

Step 1:

No matter whether $f_1\wedge f_2$ is regarded to be $:M\to T_{q_1}^0(TM)$ or $:M\to {\mathrm{\Lambda }}_{q_1}(TM)$, the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$ can be applied.

Let $V_1:M\to TM,...,V_{q_1+q_2}:M\to TM$ be any $C^{\mathrm{\infty }}$ vectors fields.

$f_1\wedge f_2(V_1,...,V_{q_1+q_2})=(q_1+q_2)!/(q_1!q_2!)Asym(f_1\otimes f_2)(V_1,...,V_{q_1+q_2})=(q_1+q_2)!/(q_1!q_2!)1/(q_1+q_2)!\sum _{\sigma }(f_1\otimes f_2)(V_{{\sigma }_1},...,V_{{\sigma }_{q_1+q_2}})=1/(q_1!q_2!)\sum _{\sigma }(f_1(V_{{\sigma }_1},...,V_{{\sigma }_{q_1}})f_2(V_{{\sigma }_{q_1+1}},...,V_{{\sigma }_{q_1+q_2}})$, but each $f_1(V_{{\sigma }_1},...,V_{{\sigma }_{q_1}})$ and $f_2(V_{{\sigma }_{q_1+1}},...,V_{{\sigma }_{q_1+q_2}})$ are $C^{\mathrm{\infty }}$, by the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

So, $f_1\wedge f_2(V_1,...,V_{q_1+q_2})$ is $C^{\mathrm{\infty }}$.

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References

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

1171: q-Form over C∞ Manifold with Boundary Is C∞ iff Operation Result on Any C∞ Vectors Fields Is C∞

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

description/proof of that $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ iff operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$

---

Topics

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

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that any $(0,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any $C^{\mathrm{\infty }}$ section along any closed subset of the base space can be extended to over the whole base space with the support contained in any open neighborhood of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that any $q$-form over any $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$.

---

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:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$q$: $\in \mathbb{N}\setminus \{0\}$
$(T_q^0(TM),M,\pi )$: $=(0,q)\text{ -tensors bundle over }M$
$({\mathrm{\Lambda }}_q(TM),M,\pi )$: $=q\text{ -covectors bundle over }M$
$f$: $:M\to T_q^0(TM)$ such that $Ran(f)\subseteq {\mathrm{\Lambda }}_q(TM)$ or $:M\to {\mathrm{\Lambda }}_q(TM)$, $\in \{\text{ the sections of }\pi \}$
//

Statements:
$f\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$⟺$
$\mathrm{\forall }{V}_1,...,V_q\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors fields over }M\}(f(V_1,...,V_q):M\to \mathbb{R}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\})$
//

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2: Proof

Whole Strategy: Step 1: when $:M\to T_q^0(TM)$, see that the proposition holds; Step 2: suppose that $:M\to {\mathrm{\Lambda }}_q(TM)$; Step 3: suppose that $f(V_1,...,V_q)\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$, and see that $f$ is $C^{\mathrm{\infty }}$, by taking an $r^{\prime }$-$r$-open-balls charts pair or an $r^{\prime }$-$r$-open-half-balls charts pair around each $m\in M$, $(U_m^{\prime }\subseteq M,{\varphi }_m^{\prime })$ and $(U_m\subseteq M,{\varphi }_m)$, and the induced chart, $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$, and seeing that the components function of $f$ is $C^{\mathrm{\infty }}$; Step 4: suppose that $f$ is $C^{\mathrm{\infty }}$, and see that $f(V_1,...,V_q)$ is $C^{\mathrm{\infty }}$ over a chart, $(U_m\subseteq M,{\varphi }_m)$, around each $m\in M$.

Step 1:

When $f$ is $:M\to T_q^0(TM)$, $f$ is really just a special type of $(0,q)$-tensors field, so, the proposition that any $(0,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$ if and only if the operation result on any $C^{\mathrm{\infty }}$ vectors fields is $C^{\mathrm{\infty }}$ applies.

Step 2:

Let us suppose that $f$ is $:M\to {\mathrm{\Lambda }}_q(TM)$.

Step 3:

Let us suppose that $f(V_1,...,V_q)\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$.

Let $m\in M$ be any.

Let us take any $r^{\prime }$-$r$-open-balls charts pair or any $r^{\prime }$-$r$-open-half-balls charts pair around $m$, $(U_m^{\prime }\subseteq M,{\varphi }_m^{\prime })$ and $(U_m\subseteq M,{\varphi }_m)$, which is possible by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$, and take the induced chart, $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$.

Let us take $V_j=V_j^{l_j}\partial /\partial x^{l_j}$ over $U_m^{\prime }$ as $V_j^{l_j^{\prime }}\equiv 1$ and $V_j^{l_j}\equiv 0$ for each $l_j\ne l_j^{\prime }$ where $l_1^{\prime }<...<l_q^{\prime }$. $V_j$ is $C^{\mathrm{\infty }}$ over $U_m^{\prime }$. $V_j$ is $C^{\mathrm{\infty }}$ over $\overline{U_m}\subseteq U_m^{\prime }$. By the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, any $C^{\mathrm{\infty }}$ section along any closed subset of the base space can be extended to over the whole base space with the support contained in any open neighborhood of the subset, $V_j$ is extended to over $M$. The extended $V_j$ equals the original $V_j$ over $\overline{U_m}$ especially over $U_m$.

Let $m^{\prime }\in U_m$ be any.

$f(m^{\prime })=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}$.

$f(m^{\prime })(V_1,...,V_q)=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })q!Asymd{x}^{j_1}\otimes ...\otimes d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })q!1/q!\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma d{x}^{j_1}\otimes ...\otimes d{x}^{j_q}(V_{{\sigma }_1}^{l_{{\sigma }_1}}\partial /\partial x^{l_{{\sigma }_1}},...,V_{{\sigma }_q}^{l_{{\sigma }_q}}\partial /\partial x^{l_{{\sigma }_q}})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma d{x}^{j_1}(V_{{\sigma }_1}^{l_{{\sigma }_1}}\partial /\partial x^{l_{{\sigma }_1}})...d{x}^{j_q}(V_{{\sigma }_q}^{l_{{\sigma }_q}}\partial /\partial x^{l_{{\sigma }_q}})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{l_{{\sigma }_1}}{\delta }_{l_{{\sigma }_1}}^{j_1}...V_{{\sigma }_q}^{l_{{\sigma }_q}}{\delta }_{l_{{\sigma }_q}}^{j_q}=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_1^{j_{{\sigma }^{-1}(1)}}...V_q^{j_{{\sigma }^{-1}(q)}}$: $V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}$ is reordered to $V_1^{m_1}...V_q^{m_q}$, then, $V_{{\sigma }_n}^{j_n}=V_s^{m_s}$, which means that ${\sigma }_n=s$, so, $n={\sigma }^{-1}(s)$.

$V_1^{j_{{\sigma }^{-1}(1)}}...V_q^{j_{{\sigma }^{-1}(q)}}$ can be nonzero only for $\sigma =id$, because $l_1^{\prime }<...<l_q^{\prime }$.

So, $=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })V_1^{j_1}...V_q^{j_q}=f_{l_1^{\prime },...,l_q^{\prime }}(m^{\prime })$.

$f(V_1,...,V_q):U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$ by the supposition, so, $f_{l_1^{\prime },...,l_q^{\prime }}:U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$.

So, each $f_{j_1,...,j_q}$ is $C^{\mathrm{\infty }}$ over $U_m$, which means that the components function of $f$ with respect to $(U_m\subseteq M,{\varphi }_m)$ and $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$, $\widetilde{{\varphi }_m}\circ f\circ {{\varphi }_m}^{-1}$ whose components are $f_{j_1,...,j_q}\circ {{\varphi }_m}^{-1}$ s, is $C^{\mathrm{\infty }}$.

So, $f$ is $C^{\mathrm{\infty }}$.

Step 4:

Let us suppose that $f$ is $C^{\mathrm{\infty }}$.

Let $m\in M$ be any.

Let us take any chart around $m$, $(U_m\subseteq M,{\varphi }_m)$, and the induced chart, $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$.

Over $U_m$, $f=f_{j_1,...,j_q}d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}$, where $f_{j_1,...,j_q}:U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$, because it is a component of $\widetilde{{\varphi }_m}\circ f\circ {{\varphi }_m}^{-1}\circ {\varphi }_m$, while the components function of $f$ with respect to $(U_m\subseteq M,{\varphi }_m)$ and $({\pi }^{-1}(U_m)\subseteq {\mathrm{\Lambda }}_q(TM),\widetilde{{\varphi }_m})$, $\widetilde{{\varphi }_m}\circ f\circ {{\varphi }_m}^{-1}$, is $C^{\mathrm{\infty }}$ and ${\varphi }_m$ is $C^{\mathrm{\infty }}$.

$V_j=V_j^{l_j}\partial /\partial x^{l_j}$, where $V_j^{l_j}:U_m\to \mathbb{R}$ is $C^{\mathrm{\infty }}$.

$f(V_1,...,V_q)=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}d{x}^{j_1}\wedge ...\wedge d{x}^{j_q}(V_1^{l_1}\partial /\partial x^{l_1},...,V_q^{l_q}\partial /\partial x^{l_q})=\sum _{j_1<...<j_q}{f}_{j_1,...,j_q}(m^{\prime })\sum _{\sigma \in P_{\{1,...,q\}}}sgn\sigma V_{{\sigma }_1}^{j_1}...V_{{\sigma }_q}^{j_q}$ as before, which is $C^{\mathrm{\infty }}$.

As $f(V_1,...,V_q)$ is $C^{\mathrm{\infty }}$ over a neighborhood of each $m\in M$, $f(V_1,...,V_q)$ is $C^{\mathrm{\infty }}$ over $M$.

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References

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