1205: Determinant of Finite-Dimensional Vectors Space Endomorphism

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definition of determinant of finite-dimensional vectors space endomorphism

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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: Note

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %structure kind name% endomorphism.
• The reader knows a definition of determinant of square matrix over ring.

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Target Context

• The reader will have a definition of determinant of finite-dimensional vectors space endomorphism.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$f$: $:V\to V$, $\in \{\text{ the }F\text{ vectors space endomorphisms }\}$
$B$: $\in \{\text{ the bases of }V\}$
$M_{f,B}$: $=\text{ the matrix for }f\text{ with respect to }B$
$\ast det(f)$: $=det(M_{f,B})$
//

Conditions:
//

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2: Note

The point is that $det(f)$ does not depend on the choice of $B$, which is the reason why this definition is well-defined: for any vector, $v\in V$, letting the components representation with respect to $B$ of $v$ be $\tilde{v}$, the components representation with respect to another basis, $B^{\prime }$, of $v$, is $U\tilde{v}$ where $U$ is an invertible matrix; the components representation with respect to $B^{\prime }$ of $f(v)$ is $U{M}_{f,B}\tilde{v}=U{M}_{f,B}{U}^{-1}U\tilde{v}$, which means that the matrix for $f$ with respect to $B^{\prime }$ is $M_{f,B^{\prime }}:=U{M}_{f,B}{U}^{-1}$; then, $det(M_{f,B^{\prime }})=det(U{M}_{f,B}{U}^{-1})=det(U)det(M_{f,B})det(U^{-1})=det(U)det(M_{f,B})det(U{)}^{-1}=det(M_{f,B})$.

So, talking about the determinant of any finite-dimensional vectors space endomorphism makes sense, without specifying any basis.

This definition makes sense because $f$ is an endomorphism, instead of just a linear map between 2 different vectors spaces: for the between-2-different-vectors-spaces case, the basis of the codomain is inevitably different from the basis of the domain (in fact, there is no canonical sameness between 2 elements in 2 different vectors spaces), and by changing the basis of the codomain, the determinant of the corresponding matrix can change: for example, make all the elements of the codomain basis half-lengthed, then, the matrix will be the double, and the determinant will be $2^d$ times where $d$ is the dimension.

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References

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

1204: Rough q-Form over C∞ Manifold with Boundary

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definition of rough $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary

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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

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ $q$-covectors bundle over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $C^{\mathrm{\infty }}$ $(p,q)$-tensors bundle over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of rough section of continuous surjection.

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Target Context

• The reader will have a definition of rough $q$-form over $C^{\mathrm{\infty }}$ manifold with boundary.

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

Conditions:
//

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2: Note

As Description says, "rough $q$-form" may mean $:M\to T_q^0(TM)$ such that $Ran(f)\subseteq {\mathrm{\Lambda }}_q(TM)$ or mean $:M\to {\mathrm{\Lambda }}_q(TM)$, whose difference should not matter in most cases: ${\mathrm{\Lambda }}_q(TM)$ is an embedded submanifold with boundary of $T_q^0(TM)$.

Usually, we need only (non-rough) forms, but we sometimes need to talk about a rough form in order to 1st introduce a may-be-rough form and then prove that it is really a non-rough form.

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References

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

1203: Rough (p,q)-Tensors Field over C∞ Manifold with Boundary

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

definition of rough $(p,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary

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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

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ $(p,q)$-tensors bundle over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of rough section of continuous surjection.

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Target Context

• The reader will have a definition of rough $(p,q)$-tensors field over $C^{\mathrm{\infty }}$ manifold with boundary.

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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:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$p$: $\in \mathbb{N}$
$q$: $\in \mathbb{N}$
$(T_q^p(TM),M,\pi )$: $=\text{ the }{C}^{\mathrm{\infty }}(p,q)\text{ -tensors bundle over }M$
$\ast f$: $:M\to T_q^p(TM)$, $\in \{\text{ the rough sections of }\pi \}$
//

Conditions:
//

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References

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

1202: Rough Vectors Field over C∞ Manifold with Boundary

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definition of rough vectors field over $C^{\mathrm{\infty }}$ manifold with boundary

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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

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Starting Context

• The reader knows a definition of tangent vectors bundle over $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of rough section of continuous surjection.

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Target Context

• The reader will have a definition of rough vectors field over $C^{\mathrm{\infty }}$ manifold with boundary.

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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:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$(TM,M,\pi )$: $=\text{ the tangent vectors bundle over }M$
$\ast V$: $:M\to TM$
//

Conditions:
$V\in \{\text{ the rough sections of }\pi \}$
//

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2: Note

As $\pi$ is a continuous surjection (in fact, $C^{\mathrm{\infty }}$), the definition is well-defined.

Usually, we need only (non-rough) vectors fields, but we sometimes need to talk about a rough vectors field in order to 1st introduce a may-be-rough vectors field and then prove that it is really a non-rough vectors field.

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References

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

1201: Rough Section of Continuous Surjection

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

definition of rough section of continuous surjection

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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
• 3: Note

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Starting Context

• The reader knows a definition of continuous map.
• The reader knows a definition of surjection.

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Target Context

• The reader will have a definition of rough section of continuous surjection.

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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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$\pi$: $:T_1\to T_2$, $\in \{\text{ the continuous surjections }\}$
$\ast s$: $:T_2\to T_1$
//

Conditions:
$\pi \circ s:T_2\to T_2=id$
//

$s$ is called "rough section of $\pi$".

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3: Note

$\pi$ needs to be surjective, because otherwise, there would be a $t\in T_2$ that would not be mapped under $\pi$ to, and then, $\pi \circ s(t)=t$ would be impossible whatever $s$ we chose, which means that $\pi \circ s=id$ would be impossible.

Usually, we need only (non-rough) sections, but we sometimes need to talk about a rough section in order to 1st introduce a may-be-rough section and then prove that it is really a non-rough section.

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References

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

1200: For Group, This Set of Subsets Constitutes Topological Basis Constituting Topological Group with This as Neighborhoods Basis at Point

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

description/proof of that for group, this set of subsets constitutes topological basis constituting topological group with this as neighborhoods basis at point

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Topics

About: topological group

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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 topological group.
• The reader knows a definition of neighborhoods basis at point on topological space.
• The reader knows a definition of inverse of subset of group.
• The reader knows a definition of union of set.
• The reader admits some criteria for any collection of open sets to be a basis.
• The reader admits the proposition that any basis of any topological space determines the topology.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of $1$, and any positive natural number, there is a symmetric neighborhood of $1$ whose power to the natural number is contained in the neighborhood.

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Target Context

• The reader will have a description and a proof of the proposition that for any group, this set of subsets constitutes a topological basis constituting a topological group with this as a neighborhoods basis at each 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:
$G$: $\in \{\text{ the groups }\}$
$B_1$: $\subseteq Pow(G)$, with the properties specified below
$B$: $={\cup }_{g\in G}g{B}_1$
$B^{\prime }$: $={\cup }_{g\in G}{B}_{1}g$
//

Statements:
(
0) $B_1\ne \mathrm{\varnothing }\wedge \mathrm{\forall }{S}_1\in B_1(1\in S_1)$
$\wedge$
1) $\mathrm{\forall }g\in G\text{ such that }g\ne 1(\mathrm{\exists }{S}_1\in B_1(g\notin S_1))$
$\wedge$
2) $\mathrm{\forall }{S}_1,S_1^{\prime }\in B_1(\mathrm{\exists }{S}_1^{″}\in B_1(S_1^{″}\subseteq S_1\cap S_1^{\prime }))$
$\wedge$
3) $\mathrm{\forall }{S}_1\in B_1(\mathrm{\exists }{S}_1^{\prime }\in B_1({S_1^{\prime }}^{-1}{S}_1^{\prime }\subseteq S_1))$
$\wedge$
4) $\mathrm{\forall }{S}_1\in B_1,\mathrm{\forall }g\in G(\mathrm{\exists }{S}_1^{\prime }\in B_1(S_1^{\prime }\subseteq g{S}_1{g}^{-1}))$
$\wedge$
5) $\mathrm{\forall }{S}_1\in B_1,\mathrm{\forall }s\in S_1(\mathrm{\exists }{S}_1^{\prime }\in B_1(S_1^{\prime }s\subseteq S_1))$
)
$⟹$
(
(
$B\in \{\text{ the topological bases for }G\}$
$\wedge$
$G\in \{\text{ the topological groups with }B\}$
$\wedge$
$\mathrm{\forall }g\in G(gB\in \{\text{ the neighborhoods bases at }g\text{ on }G\})$
)
$\wedge$
(
$B^{\prime }\in \{\text{ the topological bases for }G\}$
$\wedge$
$G\in \{\text{ the topological groups with }{B}^{\prime }\}$
$\wedge$
$\mathrm{\forall }g\in G(Bg\in \{\text{ the neighborhoods bases at }g\text{ on }G\})$
)
)
//

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2: Proof

Whole Strategy: Step 1: see that $B$ satisfies one of some criteria for any collection of open sets to be a basis; Step 2: see that $g{B}_1$ is a neighborhoods basis at $g$; Step 3: see that the group operations of $G$ are continuous; Step 4: see that $G$ is Hausdorff; Step 5: see that $B^{\prime }$ satisfies one of some criteria for any collection of open sets to be a basis; Step 6: see that $B_{1}g$ is a neighborhoods basis at $g$; Step 7: see that the group operations of $G$ are continuous; Step 8: see that $G$ is Hausdorff.

Step 1:

Let us see that $B$ satisfies Description 2 of some criteria for any collection of open sets to be a basis.

1) $G=\cup B$ (refer to the definition of union of set)?

$1\in \cup B$, because $B_1\subseteq B$, so, $\cup B_1\subseteq \cup B$, there is an $S_1\in B_1$ such that $1\in S_1$, $S_1\subseteq \cup B_1$, and $1\in S_1\subseteq \cup B_1\subseteq \cup B$.

For each $g\in G$, $g\in \cup B$, because $g{B}_1\subseteq B$, so, $\cup g{B}_1\subseteq \cup B$, as there is an $S_1\in B_1$ such that $1\in S_1$, $g\in g{S}_1\in g{B}_1$, $g{S}_1\subseteq \cup g{B}_1$, and $g\in g{S}_1\subseteq \cup g{B}_1\subseteq \cup B$.

So, 1) holds.

2) for each sets, $S_j,S_l\in B$, and each point, $g\in S_j\cap S_l$, there is a set, $S_m\in B$, such that $g\in S_m\subseteq S_j\cap S_l$?

Let $S_j\in g_j{B}_1$, $S_l\in g_l{B}_1$, and $S_m\in g_m{B}_1$.

$S_j=g_j{S}_{1,j}$ and $S_l=g_j{S}_{1,l}$ where $S_{1,j},S_{1,l}\in B_1$.

As $g\in g_j{S}_{1,j}$, ${g_j}^{-1}g\in S_{1,j}$; ${g_l}^{-1}g\in S_{1,l}$, likewise.

By 5), there is an $S_{1,j}^{\prime }\in B_1$ such that $S_{1,j}^{\prime }{g_j}^{-1}g\subseteq S_{1,j}$; there is an $S_{1,l}^{\prime }\in B_1$ such that $S_{1,l}^{\prime }{g_l}^{-1}g\subseteq S_{1,l}$, likewise.

$S_{1,j}^{\prime }\subseteq S_{1,j}{g}^{-1}{g}_j$ and $S_{1,l}^{\prime }\subseteq S_{1,l}{g}^{-1}{g}_l$.

By 4), there is an $S_{1,j}^{″}\in B_1$ such that $S_{1,j}^{″}\subseteq g^{-1}{g}_j{S}_{1,j}^{\prime }{g_j}^{-1}g$; there is an $S_{1,l}^{″}\in B_1$ such that $S_{1,l}^{″}\subseteq g^{-1}{g}_l{S}_{1,l}^{\prime }{g_l}^{-1}g$, likewise.

So, $S_{1,j}^{″}\subseteq g^{-1}{g}_j{S}_{1,j}{g}^{-1}{g}_j{g_j}^{-1}g=g^{-1}{g}_j{S}_{1,j}$, so, $g{S}_{1,j}^{″}\subseteq g{g}^{-1}{g}_j{S}_{1,j}=g_j{S}_{1,j}$; $g{S}_{1,l}^{″}\subseteq g_l{S}_{1,l}$, likewise.

By 2), there is an $S_1\in B_1$ such that $S_1\subseteq S_{1,j}^{″}\cap S_{1,l}^{″}$.

So, $g{S}_1\subseteq g(S_{1,j}^{″}\cap S_{1,l}^{″})=(g{S}_{1,j}^{″})\cap (g{S}_{1,l}^{″})\subseteq g_j{S}_{1,j}\cap g_l{S}_{1,l}=S_j\cap S_l$.

$g\in g{S}_1$, because $1\in S_1$.

So, $g{S}_1=S_m$ will do.

So, $G$ with $B$ is a topological space, by the proposition that any basis of any topological space determines the topology.

Step 2:

Let us see that $g{B}_1$ is a neighborhoods basis at $g$.

Let $N_g$ be any neighborhood of $g$.

There is a $g^{\prime }{S}_1\in B$ such that $g\in g^{\prime }{S}_1\subseteq N_g$: the point is that $g^{\prime }$ is not guaranteed to be taken to be $g$, yet.

$g^{\prime -1}g\in S_1$. So, there is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }{g}^{\prime -1}g\subseteq S_1$, by 5).

$S_1^{\prime }{g}^{\prime -1}g=g^{\prime -1}g{g}^{-1}{g}^{\prime }{S}_1^{\prime }{g}^{\prime -1}g$.

There is an $S_1^{″}\in B_1$ such that $S_1^{″}\subseteq g^{-1}{g}^{\prime }{S}_1^{\prime }{g}^{\prime -1}g$, by 4). So, $g^{\prime -1}g{S}_1^{″}\subseteq g^{\prime -1}g{g}^{-1}{g}^{\prime }{S}_1^{\prime }{g}^{\prime -1}g=S_1^{\prime }{g}^{\prime -1}g\subseteq S_1$.

So, $g^{\prime }{g}^{\prime -1}g{S}_1^{″}\subseteq g^{\prime }{S}_1$, but $g^{\prime }{g}^{\prime -1}g{S}_1^{″}=g{S}_1^{″}$, so, $g\in g{S}_1^{″}\subseteq g^{\prime }{S}_1\subseteq N_g$.

Step 3:

Let us see that the group operations of $G$ are continuous.

Let us deal with the inverse map.

Let $N_{g^{-1}}\subseteq G$ be any neighborhood of $g^{-1}$.

There is a $g^{-1}{S}_1\in B$ such that $g^{-1}\in g^{-1}{S}_1\subseteq N_{g^{-1}}$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq g^{-1}{S}_{1}g$, by 4).

So, $S_1^{\prime }{g}^{-1}\subseteq g^{-1}{S}_1$.

There is an $S_1^{″}\in B_1$ such that ${S_1^{″}}^{-1}{S}_1^{″}\subseteq S_1^{\prime }$, by 3).

${S_1^{″}}^{-1}{S}_1^{″}{g}^{-1}\subseteq S_1^{\prime }{g}^{-1}$.

But ${S_1^{″}}^{-1}{S}_1^{″}{g}^{-1}=(g{S_1^{″}}^{-1}{S}_1^{″}{)}^{-1}$ and $S_1^{″}\subseteq {S_1^{″}}^{-1}{S}_1^{″}$, so, $(g{S}_1^{″}{)}^{-1}\subseteq S_1^{\prime }{g}^{-1}\subseteq g^{-1}{S}_1\subseteq N_{g^{-1}}$.

The inverse map is a bijection and its inverse is itself, so, the inverse map is a homeomorphism.

Let us deal with the multiplication map.

Let $N_{g_1{g}_2}\subseteq G$ be any neighborhood of $g_1{g}_2$.

There is a $g_1{g}_2{S}_1\in B$ such that $g_1{g}_2\in g_1{g}_2{S}_1\subseteq N_{g_1{g}_2}$.

$g_1{g}_2{S}_1=g_1{g}_2{S}_1{g}_2^{-1}{g}_2$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq g_2{S}_1{g}_2^{-1}$, by 4), so, $g_1{S}_1^{\prime }{g}_2\subseteq g_1{g}_2{S}_1{g}_2^{-1}{g}_2=g_1{g}_2{S}_1$.

There is an $S_1^{″}\in B_1$ such that ${S_1^{″}}^{-1}{S}_1^{″}\subseteq S_1^{\prime }$, by 3), so, $g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2\subseteq g_1{S}_1^{\prime }{g}_2\subseteq g_1{g}_2{S}_1$.

But as the inverse map is a homeomorphism, ${S_1^{″}}^{-1}$ is an open neighborhood of $1$, so, there is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq {S_1^{″}}^{-1}$, so, $g_1{S}_1^{‴}{S}_1^{″}{g}_2\subseteq g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2$.

$g_1{S}_1^{‴}{S}_1^{″}{g}_2=g_1{S}_1^{‴}{g}_2{g}_2^{-1}{S}_1^{″}{g}_2$.

There is an $S_1^{⁗}\in B_1$ such that $S_1^{⁗}\subseteq g_2^{-1}{S}_1^{″}{g}_2$, by 4), so, $g_1{S}_1^{‴}{g}_2{S}_1^{⁗}\subseteq g_1{S}_1^{‴}{S}_1^{″}{g}_2$.

So, $g_1{S}_1^{‴}{g}_2{S}_1^{⁗}\subseteq g_1{S}_1^{‴}{S}_1^{″}{g}_2\subseteq g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2\subseteq g_1{g}_2{S}_1\subseteq N_{g_1{g}_2}$.

Step 4:

Let us see that $G$ is Hausdorff.

Let $g_1,g_2\in G$ be any such that $g_1\ne g_2$.

${g_1}^{-1}{g}_2\ne 1$, so, there is an $S_1\in B_1$ such that ${g_1}^{-1}{g}_2\notin S_1$, by 1).

There is a symmetric neighborhood of $1$, $N_1\subseteq G$, such that $N_1^2\subseteq S_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of $1$, and any positive natural number, there is a symmetric neighborhood of $1$ whose power to the natural number is contained in the neighborhood.

Let us see that $N_1\cap {g_1}^{-1}{g}_2{N}_1=\mathrm{\varnothing }$.

Let us suppose that $N_1\cap {g_1}^{-1}{g}_2{N}_1\ne \mathrm{\varnothing }$.

There would be an $n\in N_1\cap {g_1}^{-1}{g}_2{N}_1$.

$n={g_1}^{-1}{g}_2{n}^{\prime }$ for an $n^{\prime }\in N_1$.

${g_1}^{-1}{g}_2=n{n}^{\prime -1}$. But as $N_1$ is symmetric, $n^{\prime -1}\in N_1^{-1}=N_1$, so, ${g_1}^{-1}{g}_2\in N_1^2\subseteq S_1$, a contradiction against ${g_1}^{-1}{g}_2\notin S_1$.

So, $N_1\cap {g_1}^{-1}{g}_2{N}_1=\mathrm{\varnothing }$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq N_1$, and $S_1^{\prime }\cap {g_1}^{-1}{g}_2{S}_1^{\prime }\subseteq N_1\cap {g_1}^{-1}{g}_2{N}_1=\mathrm{\varnothing }$.

So, $g_1(S_1^{\prime }\cap {g_1}^{-1}{g}_2{S}_1^{\prime })=g_1{S}_1^{\prime }\cap g_2{S}_1^{\prime }=\mathrm{\varnothing }$.

Step 5:

Let us see that $B^{\prime }$ satisfies Description 2 of some criteria for any collection of open sets to be a basis.

1) $G=\cup B$ (refer to the definition of union of set)?

$1\in \cup B^{\prime }$, because $B_1\subseteq B^{\prime }$, so, $\cup B_1\subseteq \cup B^{\prime }$, there is an $S_1\in B_1$ such that $1\in S_1$, $S_1\subseteq \cup B_1$, and $1\in S_1\subseteq \cup B_1\subseteq \cup B^{\prime }$.

For each $g\in G$, $g\in \cup B^{\prime }$, because $B_{1}g\subseteq B^{\prime }$, so, $\cup B_{1}g\subseteq \cup B^{\prime }$, as there is an $S_1\in B_1$ such that $1\in S_1$, $g\in S_{1}g\in B_{1}g$, $S_{1}g\subseteq \cup B_{1}g$, and $g\in S_{1}g\subseteq \cup B_{1}g\subseteq \cup B^{\prime }$.

So, 1) holds.

2) for each sets, $S_j,S_l\in B^{\prime }$, and each point, $g\in S_j\cap S_l$, there is a set, $S_m\in B^{\prime }$, such that $g\in S_m\subseteq S_j\cap S_l$?

Let $S_j\in B_1{g}_j$, $S_l\in B_1{g}_l$, and $S_m\in B_1{g}_m$.

$S_j=S_{1,j}{g}_j$ and $S_l=S_{1,l}{g}_j$ where $S_{1,j},S_{1,l}\in B_1$.

As $g\in S_{1,j}{g}_j$, $g{g_j}^{-1}\in S_{1,j}$; $g{g_l}^{-1}\in S_{1,l}$, likewise.

By 5), there is an $S_{1,j}^{\prime }\in B_1$ such that $S_{1,j}^{\prime }g{g_j}^{-1}\subseteq S_{1,j}$; there is an $S_{1,l}^{\prime }\in B_1$ such that $S_{1,l}^{\prime }g{g_l}^{-1}\subseteq S_{1,l}$, likewise.

$S_{1,j}^{\prime }\subseteq S_{1,j}{g}_j{g}^{-1}$ and $S_{1,l}^{\prime }\subseteq S_{1,l}{g}_l{g}^{-1}$.

So, $S_{1,j}^{\prime }g\subseteq S_{1,j}{g}_j{g}^{-1}g=S_{1,j}{g}_j$; $S_{1,l}^{\prime }g\subseteq S_{1,l}{g}_l$, likewise.

By 2), there is an $S_1\in B_1$ such that $S_1\subseteq S_{1,j}^{\prime }\cap S_{1,l}^{\prime }$.

So, $S_{1}g\subseteq (S_{1,j}^{\prime }\cap S_{1,l}^{\prime })g=(S_{1,j}^{\prime }g)\cap (S_{1,l}^{\prime }g)\subseteq (S_{1,j}{g}_j)\cap (S_{1,l}{g}_l)=S_j\cap S_l$.

$g\in S_{1}g$, because $1\in S_1$.

So, $S_{1}g=S_m$ will do.

So, $G$ with $B^{\prime }$ is a topological space, by the proposition that any basis of any topological space determines the topology.

Step 6:

Let us see that $B_{1}g$ is a neighborhoods basis at $g$.

Let $N_g$ be any neighborhood of $g$.

There is a $S_1{g}^{\prime }\in B^{\prime }$ such that $g\in S_1{g}^{\prime }\subseteq N_g$: the point is that $g^{\prime }$ is not guaranteed to be taken to be $g$, yet.

$g{g}^{\prime -1}\in S_1$. So, there is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }g{g}^{\prime -1}\subseteq S_1$, by 5).

So, $S_1^{\prime }g{g}^{\prime -1}{g}^{\prime }\subseteq S_1{g}^{\prime }$, but $S_1^{\prime }g{g}^{\prime -1}{g}^{\prime }=S_1^{\prime }g$, so, $g\in S_1^{\prime }g\subseteq S_1{g}^{\prime }\subseteq N_g$.

Step 7:

Let us see that the group operations of $G$ are continuous.

Let us deal with the inverse map.

Let $N_{g^{-1}}\subseteq G$ be any neighborhood of $g^{-1}$.

There is a $S_1{g}^{-1}\in B^{\prime }$ such that $g^{-1}\in S_1{g}^{-1}\subseteq N_{g^{-1}}$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq g{S}_1{g}^{-1}$, by 4).

So, $g^{-1}{S}_1^{\prime }\subseteq S_1{g}^{-1}$.

There is an $S_1^{″}\in B_1$ such that ${S_1^{″}}^{-1}{S}_1^{″}\subseteq S_1^{\prime }$, by 3).

$g^{-1}{S_1^{″}}^{-1}{S}_1^{″}\subseteq g^{-1}{S}_1^{\prime }$.

But $g^{-1}{S_1^{″}}^{-1}{S}_1^{″}=({S_1^{″}}^{-1}{S}_1^{″}g{)}^{-1}$ and $S_1^{″}\subseteq {S_1^{″}}^{-1}{S}_1^{″}$, so, $(S_1^{″}g{)}^{-1}\subseteq g^{-1}{S}_1^{\prime }\subseteq S_1{g}^{-1}\subseteq N_{g^{-1}}$.

The inverse map is a bijection and its inverse is itself, so, the inverse map is a homeomorphism.

Let us deal with the multiplication map.

Let $N_{g_1{g}_2}\subseteq G$ be any neighborhood of $g_1{g}_2$.

There is a $S_1{g}_1{g}_2\in B^{\prime }$ such that $g_1{g}_2\in S_1{g}_1{g}_2\subseteq N_{g_1{g}_2}$.

$S_1{g}_1{g}_2=g_1{g_1}^{-1}{S}_1{g}_1{g}_2$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq {g_1}^{-1}{S}_1{g}_1$, by 4), so, $g_1{S}_1^{\prime }{g}_2\subseteq g_1{g_1}^{-1}{S}_1{g}_1{g}_2=S_1{g}_1{g}_2$.

There is an $S_1^{″}\in B_1$ such that ${S_1^{″}}^{-1}{S}_1^{″}\subseteq S_1^{\prime }$, by 3), so, $g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2\subseteq g_1{S}_1^{\prime }{g}_2\subseteq S_1{g}_1{g}_2$.

But as the inverse map is a homeomorphism, ${S_1^{″}}^{-1}$ is an open neighborhood of $1$, so, there is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq {S_1^{″}}^{-1}$, so, $g_1{S}_1^{‴}{S}_1^{″}{g}_2\subseteq g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2$.

$g_1{S}_1^{‴}{S}_1^{″}{g}_2=g_1{S}_1^{‴}{g}_1^{-1}{g}_1{S}_1^{″}{g}_2$.

There is an $S_1^{⁗}\in B_1$ such that $S_1^{⁗}\subseteq g_1{S}_1^{‴}{g}_1^{-1}$, by 4), so, $S_1^{⁗}{g}_1{S}_1^{″}{g}_2\subseteq g_1{S}_1^{‴}{S}_1^{″}{g}_2$.

So, $S_1^{⁗}{g}_1{S}_1^{″}{g}_2\subseteq g_1{S}_1^{‴}{S}_1^{″}{g}_2\subseteq g_1{S_1^{″}}^{-1}{S}_1^{″}{g}_2\subseteq S_1{g}_1{g}_2\subseteq N_{g_1{g}_2}$.

Step 8:

Let us see that $G$ is Hausdorff.

Let $g_1,g_2\in G$ be any such that $g_1\ne g_2$.

$g_1{g_2}^{-1}\ne 1$, so, there is an $S_1\in B_1$ such that $g_1{g_2}^{-1}\notin S_1$, by 1).

There is a symmetric neighborhood of $1$, $N_1\subseteq G$, such that $N_1^2\subseteq S_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of $1$, and any positive natural number, there is a symmetric neighborhood of $1$ whose power to the natural number is contained in the neighborhood.

Let us see that $N_1\cap N_1{g}_1{g_2}^{-1}=\mathrm{\varnothing }$.

Let us suppose that $N_1\cap N_1{g}_1{g_2}^{-1}\ne \mathrm{\varnothing }$.

There would be an $n\in N_1\cap N_1{g}_1{g_2}^{-1}$.

$n=n^{\prime }{g}_1{g_2}^{-1}$ for an $n^{\prime }\in N_1$.

$g_1{g_2}^{-1}=n^{\prime -1}n$. But as $N_1$ is symmetric, $n^{\prime -1}\in N_1^{-1}=N_1$, so, $g_1{g_2}^{-1}\in N_1^2\subseteq S_1$, a contradiction against $g_1{g_2}^{-1}\notin S_1$.

So, $N_1\cap N_1{g}_1{g_2}^{-1}=\mathrm{\varnothing }$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq N_1$, and $S_1^{\prime }\cap S_1^{\prime }{g}_1{g_2}^{-1}\subseteq N_1\cap N_1{g}_1{g_2}^{-1}=\mathrm{\varnothing }$.

So, $(S_1^{\prime }\cap S_1^{\prime }{g}_1{g_2}^{-1})g_2=S_1^{\prime }{g}_2\cap S_1^{\prime }{g}_1{g_2}^{-1}{g}_2=S_1^{\prime }{g}_2\cap S_1^{\prime }{g}_1=\mathrm{\varnothing }$.

---

References

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

1199: For Topological Group, Closed Subset Is Intersection of Subset Multiplied by Elements of Neighborhoods Basis at 1

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description/proof of that for topological group, closed subset is intersection of subset multiplied by elements of neighborhoods basis at $1$

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Topics

About: topological group

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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 topological group.
• The reader knows a definition of neighborhoods basis at point on topological space.
• The reader knows a definition of closed subset of topological space.
• The reader admits the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.
• 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 topological group, any closed subset is the intersection of the subset multiplied by the elements of any neighborhoods basis at $1$.

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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:
$G$: $\in \{\text{ the topological groups }\}$
$B_1$: $\in \{\text{ the neighborhoods bases at }1\text{ on }G\}$
$C$: $\in \{\text{ the closed subsets of }G\}$
//

Statements:
$C={\cap }_{S_1\in B_1}{S}_{1}C={\cap }_{S_1\in B_1}C{S}_1$
//

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2: Proof

Whole Strategy: Step 1: see that $C\subseteq {\cap }_{S_1\in B_1}{S}_{1}C$; Step 2: see that ${\cap }_{S_1\in B_1}{S}_{1}C\subseteq C$, by seeing that for each $g\in {\cap }_{S_1\in B_1}{S}_{1}C$, $g\in C$ or $g$ is an accumulation point of $C$; Step 3: see that $C\subseteq {\cap }_{S_1\in B_1}C{S}_1$; Step 4: see that ${\cap }_{S_1\in B_1}C{S}_1\subseteq C$, by seeing that for each $g\in {\cap }_{S_1\in B_1}C{S}_1$, $g\in C$ or $g$ is an accumulation point of $C$.

Step 1:

Let us see that $C\subseteq {\cap }_{S_1\in B_1}{S}_{1}C$.

For each $c\in C$, for each $S_1\in B_1$, $c\in S_{1}C$, because $1\in S_1$, so, $c\in {\cap }_{S_1\in B_1}{S}_{1}C$.

Step 2:

Let us see that ${\cap }_{S_1\in B_1}{S}_{1}C\subseteq C$.

Let $g\in {\cap }_{S_1\in B_1}{S}_{1}C$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

As $g{B}_1$ is a neighborhood basis at $g$, by the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point, there is an $S_1\in B_1$ such that $g{S}_1\subseteq N_g$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq g{S}_1{g}^{-1}$, by 4) in the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point.

There is a symmetric neighborhood of $1$, $S_1^{″}\subseteq G$, such that $S_1^{″}\subseteq S_1^{\prime }$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

So, $S_1^{″}\subseteq g{S}_1{g}^{-1}$, so, $S_1^{″}g\subseteq g{S}_1{g}^{-1}g=g{S}_1$.

There is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq S_1^{″}$.

$g\in S_1^{‴}C\subseteq S_1^{″}C$, because $g\in {\cap }_{S_1\in B_1}{S}_{1}C$.

That means that $g=sc$ where $s\in S_1^{″}$ and $c\in C$.

As $S_1^{″}g\subseteq g{S}_1$, $S_1^{″}sc\subseteq g{S}_1$, but as $S_1^{″}$ is symmetric, $s^{-1}\in S_1^{″}$, so, $c=s^{-1}sc\in g{S}_1\subseteq N_g$, which means that $g\in C$ or $g$ is an accumulation point of $C$, which means that $g$ is in the closure of $C$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, $g\in \bar{C}$.

But as $C$ is closed, $\bar{C}=C$, so, $g\in \bar{C}=C$.

Step 3:

Let us see that $C\subseteq {\cap }_{S_1\in B_1}C{S}_1$.

As is expected, the logic is parallel to Step 1.

For each $c\in C$, for each $S_1\in B_1$, $c\in C{S}_1$, because $1\in S_1$, so, $c\in {\cap }_{S_1\in B_1}C{S}_1$.

Step 4:

Let us see that ${\cap }_{S_1\in B_1}C{S}_1\subseteq C$.

As is expected, the logic is parallel to Step 2, in fact, this is simpler: $S_1^{\prime }$ is not necessary as it is introduced in order to deal with the $S_{1}C$ order.

Let $g\in {\cap }_{S_1\in B_1}C{S}_1$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

As $g{B}_1$ is a neighborhood basis at $g$, by the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point, there is an $S_1\in B_1$ such that $g{S}_1\subseteq N_g$.

There is a symmetric neighborhood of $1$, $S_1^{″}\subseteq G$, such that $S_1^{″}\subseteq S_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

There is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq S_1^{″}$.

$g\in C{S}_1^{‴}\subseteq C{S}_1^{″}$, because $g\in {\cap }_{S_1\in B_1}C{S}_1$.

That means that $g=cs$ where $s\in S_1^{″}$ and $c\in C$.

As $S_1^{″}\subseteq S_1$, $g{S}_1^{″}=cs{S}_1^{″}\subseteq g{S}_1$, but as $S_1^{″}$ is symmetric, $s^{-1}\in S_1^{″}$, so, $c=cs{s}^{-1}\in g{S}_1\subseteq N_g$, which means that $g\in C$ or $g$ is an accumulation point of $C$, which means that $g$ is in the closure of $C$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, $g\in \bar{C}$.

But as $C$ is closed, $\bar{C}=C$, so, $g\in \bar{C}=C$.

---

References

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

1198: For Topological Group, Neighborhoods Basis at 1 Satisfies These Properties and Point Multiplied by Neighborhoods Basis at 1 Is Neighborhoods Basis at Point

<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 group, neighborhoods basis at $1$ satisfies these properties and point multiplied by neighborhoods basis at $1$ is neighborhoods basis at point

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Topics

About: topological group

---

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 topological group.
• The reader knows a definition of neighborhoods basis at point on topological space.
• The reader knows a definition of inverse of subset of group.
• The reader admits the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of $1$ such that the element multiplied from left by the neighborhood of $1$ and multiplied from right by the inverse of the neighborhood of $1$ is contained in the neighborhood of the element.
• The reader admits the proposition that for any group, the inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the 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:
$G$: $\in \{\text{ the topological groups }\}$
$B_1$: $\in \{\text{ the neighborhoods bases at }1\text{ on }G\}$
//

Statements:
1) $\mathrm{\forall }g\in G\text{ such that }g\ne 1(\mathrm{\exists }{N}_1\in B_1(g\notin N_1))$
$\wedge$
2) $\mathrm{\forall }{N}_1,N_1^{\prime }\in B_1(\mathrm{\exists }{N}_1^{″}\in B_1(N_1^{″}\subseteq N_1\cap N_1^{\prime }))$
$\wedge$
3) $\mathrm{\forall }{N}_1\in B_1(\mathrm{\exists }{N}_1^{\prime }\in B_1({N_1^{\prime }}^{-1}{N}_1^{\prime }\subseteq N_1))$
$\wedge$
4) $\mathrm{\forall }{N}_1\in B_1,\mathrm{\forall }g\in G(\mathrm{\exists }{N}_1^{\prime }\in B_1(N_1^{\prime }\subseteq g{N}_1{g}^{-1}))$
$\wedge$
5) $\mathrm{\forall }{N}_1\in B_1,\mathrm{\forall }n\in N_1\text{ such that }\mathrm{\exists }{U}_n\in \{\text{ the open neighborhoods of }n\}(n\in U_n\subseteq N_1)(\mathrm{\exists }{N}_1^{\prime }\in B_1(N_1^{\prime }n\subseteq N_1))$
$\wedge$
$\mathrm{\forall }g\in G(g{B}_1\in \{\text{ the neighborhoods bases at }g\text{ on }G\})$
$\wedge$
$\mathrm{\forall }g\in G(B_{1}g\in \{\text{ the neighborhoods bases at }g\text{ on }G\})$
//

$g{B}_1$ means $\{g{S}_1|S_1\in B_1\}$; $B_{1}g$ means $\{S_{1}g|S_1\in B_1\}$.

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2: Proof

Whole Strategy: use the Hausdorff-ness and the continuousness of the operations; Step 1: see that 1) holds; Step 2: see that 2) holds; Step 3: see that 3) holds; Step 4: see that 4) holds; Step 5: see that 5) holds; Step 6: see that $g{B}_1$ is a neighborhoods basis at $g$; Step 7: see that $B_{1}g$ is a neighborhoods basis at $g$.

Step 1:

Let us see that 1) holds.

As $G$ is Hausdorff, there is an open neighborhood of $1$, $U_1\subseteq G$, and an open neighborhood of $g$, $U_g\subseteq G$, such that $U_1\cap U_g=\mathrm{\varnothing }$.

There is a $N_1\in B_1$ such that $N_1\subseteq U_1$, by the definition of neighborhoods basis at point on topological space.

$N_1\cap U_g=\mathrm{\varnothing }$, so, $g\notin N_1$.

Step 2:

Let us see that 2) holds.

$N_1\cap N_1^{\prime }$ is a neighborhood of $1$, by the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.

So, there is a $N_1^{″}\in B_1$ such that $N_1^{″}\subseteq N_1\cap N_1^{\prime }$, by the definition of neighborhoods basis at point on topological space.

Step 3:

Let us see that 3) holds.

There is a (symmetric) neighborhood of $1$, $N_1^{″}\subseteq G$, such that ${N_1^{″}}^{-1}1N_1^{″}\subseteq N_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of $1$ such that the element multiplied from left by the neighborhood of $1$ and multiplied from right by the inverse of the neighborhood of $1$ is contained in the neighborhood of the element: $g$ in the proposition is taken to be $1$ and ${N_1^{″}}^{-1}1N_1^{″}=N_1^{″}1{N_1^{″}}^{-1}$ because $N_1^{″}$ is symmetric.

But ${N_1^{″}}^{-1}1N_1^{″}={N_1^{″}}^{-1}{N}_1^{″}$.

There is an $N_1^{\prime }\in B_1$ such that $N_1^{\prime }\subseteq N_1^{″}$, by the definition of neighborhoods basis at point on topological space.

${N_1^{\prime }}^{-1}\subseteq {N_1^{″}}^{-1}$, by the proposition that for any group, the inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset.

So, ${N_1^{\prime }}^{-1}{N}_1^{\prime }\subseteq {N_1^{″}}^{-1}{N}_1^{″}$.

So, ${N_1^{\prime }}^{-1}{N}_1^{\prime }\subseteq N_1$.

Step 4:

Let us see that 4) holds.

The conjugation map by $g$, $f:G\to G,g^{\prime }\mapsto g{g}^{\prime }{g}^{-1}$, is a homeomorphism, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.

$1\in g{N}_1{g}^{-1}$, because $1\in N_1$ and $g1g^{-1}=1$.

So, $g{N}_1{g}^{-1}$ is a neighborhood of $1$: while $N_1$ contains an open neighborhood of $1$, $U_1$, $g{N}_1{g}^{-1}=f(N_1)$ contains $f(U_1)$, which is an open neighborhood of $1$: $f(1)=1$.

So, there is an $N_1^{\prime }\in B_1$ such that $N_1^{\prime }\subseteq g{N}_1{g}^{-1}$, by the definition of neighborhoods basis at point on topological space.

Step 5:

Let us see that 5) holds.

The multiplication-by-element-from-right map by $n^{-1}$, $f:G\to G,g^{\prime }\mapsto g^{\prime }{n}^{-1}$, is a homeomorphism, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.

$1\in U_n{n}^{-1}\subseteq N_1{n}^{-1}$, because $n\in U_n$ and $n{n}^{-1}=1$.

So, $N_1{n}^{-1}$ is a neighborhood of $1$: $N_1{n}^{-1}=f(N_1)$ contains $U_n{n}^{-1}=f(U_n)$, which is an open neighborhood of $1$.

So, there is an $N_1^{\prime }\in B_1$ such that $N_1^{\prime }\subseteq N_1{n}^{-1}$, by the definition of neighborhoods basis at point on topological space.

So, $N_1^{\prime }n\subseteq N_1{n}^{-1}n=N_1$.

Step 6:

Let $g\in G$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

$g^{-1}{N}_g$ is a neighborhood of $1$, because while the multiplication-by-$g^{-1}$-from-left-map is a homeomorphism by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, an open neighborhood of $g$ contained in $N_g$ is mapped into $g^{-1}{N}_g$ as an open neighborhood of $1$.

So, there is an $S_1\in B_1$ such that $1\in S_1\subseteq g^{-1}{N}_g$.

So, $g\in g{S}_1\subseteq g{g}^{-1}{N}_g=N_g$, while $g{S}_1$ is a neighborhood of $g$, because while the multiplication-by-$g$-from-left-map is a homeomorphism by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, an open neighborhood of $1$ contained in $S_1$ is mapped into $g{S}_1$ as an open neighborhood of $g$.

Step 7:

Let $g\in G$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

$N_g{g}^{-1}$ is a neighborhood of $1$, because while the multiplication-by-$g^{-1}$-from-right-map is a homeomorphism by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, an open neighborhood of $g$ contained in $N_g$ is mapped into $N_g{g}^{-1}$ as an open neighborhood of $1$.

So, there is an $S_1\in B_1$ such that $1\in S_1\subseteq N_g{g}^{-1}$.

So, $g\in S_{1}g\subseteq N_g{g}^{-1}g=N_g$, while $S_{1}g$ is a neighborhood of $g$, because while the multiplication-by-$g$-from-right-map is a homeomorphism by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, an open neighborhood of $1$ contained in $S_1$ is mapped into $S_{1}g$ as an open neighborhood of $g$.

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References

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

1197: For Group with Topology with Continuous Operations (Especially, Topological Group), Element, and Neighborhood of Element, There Is Symmetric Neighborhood of 1 s.t. Element Multiplied from Left by Neighborhood of 1 and Multiplied from Right by Inverse of Neighborhood of 1 Is Contained in Neighborhood of Element

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

description/proof of that for group with topology with continuous operations (especially, topological group), element, and neighborhood of element, there is symmetric neighborhood of $1$ s.t. element multiplied from left by neighborhood of $1$ and multiplied from right by inverse of neighborhood of $1$ is contained in neighborhood of element

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Topics

About: group
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 neighborhood of point on topological space.
• The reader knows a definition of symmetric subset of group.
• The reader knows a definition of product topology.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
• The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
• The reader admits the proposition that for any topological group, the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.
• The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.

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Target Context

• The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of $1$ such that the element multiplied from left by the neighborhood of $1$ and multiplied from right by the inverse of the neighborhood of $1$ is contained in the neighborhood of the element.

---

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:
$G$: $\in \{\text{ the groups }\}$ with any topology such that the group operations are continuous
$g$: $\in G$
$N_g$: $\in \{\text{ the neighborhoods of }g\text{ on }G\}$
//

Statements:
$\mathrm{\exists }{N}_1\in \{\text{ the symmetric neighborhoods of }1\text{ on }G\}(N_{1}g{N_1}^{-1}\subseteq N_g)$
//

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2: Proof

Whole Strategy: Step 1: think of the continuous $f:G\times G\to G,(g_1,g_2)\mapsto g_{1}g{g}_2$ and see that $f((1,1))=g$; Step 2: take an open neighborhood of $(1,1)$, $U_{(1,1)}$, such that $f(U_{(1,1)})\subseteq N_g$ and take an open neighborhood of $1$, $U_1$, such that $U_1\times U_1\subseteq U_{(1,1)}$; Step 3: take a symmetric neighborhood of $1$, $N_1\subseteq U_1$, and see that $f(N_1\times N_1)\subseteq N_g$ and $f(N_1\times N_1)=N_{1}g{N_1}^{-1}$.

Step 1:

Let us think of $f:G\times G\to G,(g_1,g_2)\mapsto g_{1}g{g}_2$.

$f$ is the composition of the maps, $:G\times G\to G\times G,(g_1,g_2)\mapsto (g_1,g{g}_2)$ and $:G\times G\to G,(g_1,g_2)\mapsto g_1{g}_2$.

The former is continuous, because $:g_2\mapsto g{g}_2$ is continuous, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, and so $:G\times G\to G\times G,(g_1,g_2)\mapsto (g_1,g{g}_2)$ is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.

The latter is continuous, because it is the multiplication operation.

So, $f$ is continuous as the composition of the continuous maps, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.

$f((1,1))=1g1=g$.

Step 2:

As $f$ is continuous, there is an open neighborhood of $(1,1)$, $U_{(1,1)}\subseteq G\times G$, such that $f(U_{(1,1)})\subseteq N_g$.

By the definition of product topology, there is an open neighborhood of $1$, $U_1\subseteq G$, such that $U_1\times U_1\subseteq U_{(1,1)}$: while there are some open neighborhoods of $1$, ${U_1}^{\prime },{U_1}^{″}\subseteq G$, such that ${U_1}^{\prime }\times {U_1}^{″}\subseteq U_{(1,1)}$, we can take $U_1:={U_1}^{\prime }\cap {U_1}^{″}$.

Step 3:

There is a symmetric neighborhood of $1$, $N_1\subseteq G$, such that $N_1\subseteq U_1$, by the proposition that for any topological group, the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

As $N_1\times N_1\subseteq U_{(1,1)}$, $f(N_1\times N_1)\subseteq f(U_{(1,1)})\subseteq N_g$.

Let us see that $f(N_1\times N_1)=N_{1}g{N}_1$.

For each $g^{\prime }\in f(N_1\times N_1)$, $g^{\prime }=g_{1}g{g}_2$ where $g_1,g_2\in N_1$, so, $g^{\prime }\in N_{1}g{N}_1$; for each $g^{\prime }\in N_{1}g{N}_1$, $g^{\prime }=g_{1}g{g}_2$ where $g_1,g_2\in N_1$, but $g_{1}g{g}_2=f(g_1,g_2)$, so, $g^{\prime }\in f(N_1\times N_1)$.

But as $N_1$ is symmetric, $N_{1}g{N}_1=N_{1}g{N_1}^{-1}$.

So, $N_{1}g{N_1}^{-1}=f(N_1\times N_1)\subseteq N_g$.

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References

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