nyuuzyou/wikis huggingface dataset:
Dihedral Group D6 is Internal Direct Product of C2 with D3

== Theorem ==
The dihedral group $D_6$ is an internal direct product of the cyclic group $C_2$ of order $2$ and the dihedral group $D_3$:
:$D_6 = C_2 \times D_3$
== Proof ==
Let $G$ be defined by its group presentation:
:$G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^{-1} }$
or:
:$G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^5}$
Let $z$ denote $x^3$.
Then:
So $z$ commutes with $y$.
As $z$ is a power of $x$, $z$ also commutes with $x$.
Hence by definition of center:
:$z \in \map Z G$
It follows that $\gen z$ is a normal subgroup of order $2$.
Let $N$ be the subgroup of $G$ generated by $x^2$ and $y$.
Note that:
Hence the generator of $N$ satisfies:
:$\paren{x^2}^3 = e = y^2$
and:
:$y x^2 y^{-1} = x^{-2}$
Let $w := x^3$.
Then $N$ is generated by $w$ and $y$ where:
:$w^3 = 1 = y^2$
and:
:$w y = y w^2 = y w^{-1}$
and it is seen that $N$ is isomorphic to the dihedral group $D_3$.
It is now to be shown that $G$ is an internal group direct product of $N$ and $\gen z$.
We have that $N \cap \gen z = \set e$.
From the Internal Direct Product Theorem, we need to prove only that $N$ is a normal subgroup of $G$.
We have that:
:$\order N = 6$
where $\order N$ denotes the order of $N$.
We also have that $N$ is a subgroup of its normalizer $\map {N_G} N$.
Hence by Lagrange's Theorem:
:$6 \divides \order {\map {N_G} N}$
where $\divides$ denotes divisibility.
Again by Lagrange's Theorem:
:$\order {\map {N_G} N} \divides 12$
We have:
demonstrating that $x$ is conjugate to $w$.
Then:
demonstrating that $x$ is conjugate to $y$.
Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 6$.
As $6 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 12$, it follows that:
:$\map {N_G} N = G$
and so $N$ is normal in $G$.
Thus:
: $N$ and $\gen z$ are normal in $G$
: $N \cap \gen z = \set e$
: $N \gen z = G$
and it therefore follows from the Internal Direct Product Theorem that:
:$G = C_2 \times D_3$
== Sources ==
* : Chapter $13$: Direct products: Exercise $3$
Category:Dihedral Group D6
Category:Examples of Internal Group Direct Products
Dihedral Group is Group

== Theorem ==
Let $D_n$ be the dihedral group of order $2 n$.
Then $D_n$ is indeed a group.
== Proof ==
$D_n$ is by definition the symmetry group of the regular $n$-gon.
The result follows from Symmetry Group is Group.
== Sources ==
* : $\S 1.8$
Category:Dihedral Groups
Dihedral Group is Non-Abelian

== Theorem ==
Let $n \in \N$ be a natural number such that $n > 2$.
Let $D_n$ denote the dihedral group of order $2 n$.
Then $D_n$ is not abelian.
== Proof ==
From Group of Order less than 6 is Abelian we have that $D_1$ and $D_2$ are abelian, which is why the condition on $n$.
From Group Presentation of Dihedral Group we have:
:$\beta \alpha = \alpha^{n - 1} \beta$
for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$.
We also have:
:$\alpha^n = e$
But if $D_n$ were abelian, that would mean:
:$\beta \alpha = \beta \alpha^{n - 1}$
and so by the Cancellation Laws:
:$\alpha = \alpha^{n - 1}$
and so:
:$\alpha^2 = e$
But by definition of the dihedral group, this is not the case.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 29 \gamma$
Category:Dihedral Groups
Dilation Mapping on Topological Vector Space is Continuous

== Theorem ==
Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field.
Let $\struct {X, +_X, \circ_X, \tau_X}$ be a vector space over $K$. 
Let $\lambda \in K$. 
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is continuous.
== Proof ==
Let $f: K \times X \to X$ be the map defined by:
:$\tuple {\lambda, x} \mapsto \lambda \circ_X x$
for any $x \in X$
From the definition of a topological vector space, the mapping $f$ is continuous.
We are given that $c_\lambda$ is the dilation by $\lambda$ mapping.
That is, $c_\lambda : X \to X$ is the map:
:$\forall x \in X: \map {c_\lambda} x = \lambda \circ_X x$
So $c_\lambda$ is by definition a $\lambda$-vertical section of $f$ such that:
:$\forall x \in X: \map {c_\lambda} x = \map f {\lambda, x}$
The result follows from Vertical Section of Continuous Function is Continuous.
Category:Dilation Mappings
Category:Topological Vector Spaces
Dilation Mapping on Topological Vector Space is Homeomorphism

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $\lambda \in K \setminus \set {0_K}$. 
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is a homeomorphism.
== Proof ==
From Dilation Mapping on Topological Vector Space is Continuous, both $c_{\lambda}$ and $c_{1/\lambda}$ are continuous.
It is therefore sufficient to establish that $c_{1/\lambda}$ is the inverse mapping of $c_\lambda$.
For all $x \in X$, we have:
and:
So both $c_\lambda \circ c_{1/\lambda}$ and $c_{1/\lambda} \circ c_\lambda$ are the identity mapping for $X$.
So $c_{1/\lambda}$ is the inverse mapping of $c_\lambda$, as required.
== Sources ==
* : $1.6$: Topological vector spaces
Category:Dilation Mappings
Category:Topological Vector Spaces
Dilation of Closed Set in Topological Vector Space is Closed Set

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $F$ be a closed set in $X$. 
Let $\lambda \in K \setminus \set {0_K}$.
Then $\lambda F$ is a closed set in $X$.
== Proof 1 ==
== Proof 2 ==
Category:Dilation of Subsets of Vector Spaces
Category:Topological Vector Spaces
Category:Dilation of Closed Set in Topological Vector Space is Closed Set
Dilation of Closed Set in Topological Vector Space is Closed Set/Proof 1

== Theorem ==
== Proof ==
We aim to show that $X \setminus \paren {\lambda F}$ is open.
Since $F$ is closed, $X \setminus F$ is open.
It follows from Dilation of Open Set in Topological Vector Space is Open that $\lambda \paren {X \setminus F}$ is open.
From Dilation of Complement of Set in Vector Space, we have: 
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$.
Since we have established that $\lambda \paren {X \setminus F}$ is open, it follows that $X \setminus \paren {\lambda F}$ is open.
So $\lambda F$ is closed.
Category:Dilation of Closed Set in Topological Vector Space is Closed Set
Dilation of Closed Set in Topological Vector Space is Closed Set/Proof 2

== Theorem ==
== Proof ==
Define a mapping $c_\lambda : X \to X$ by:
:$\map {c_\lambda} x = \lambda x$
for each $x \in X$.
From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.
From Definition 4 of a homeomorphism, $c_\lambda$ is therefore a closed mapping.
Hence $c_\lambda \sqbrk F = \lambda F$ is closed.
Category:Dilation of Closed Set in Topological Vector Space is Closed Set
Dilation of Closure of Set in Topological Vector Space is Closure of Dilation

== Theorem ==
Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$. 
Let $A \subseteq X$.
Let $\lambda \in F \setminus \set {0_F}$.
Then we have: 
:$\lambda A^- = \paren {\lambda A}^-$
where $A^-$ denotes the closure of $A$. 
== Proof 1 ==
== Proof 2 ==
Category:Topological Vector Spaces
Category:Set Closures
Category:Dilation of Subsets of Vector Spaces
Category:Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
Dilation of Closure of Set in Topological Vector Space is Closure of Dilation/Proof 1

== Theorem ==
== Proof ==
From Set Closure as Intersection of Closed Sets, we have: 
:$\ds A^- = \bigcap \leftset {K \supseteq A: K}$ is closed in $\rightset X$
and:
:$\ds \paren {\lambda A}^- = \bigcap \leftset {K \supseteq \lambda A: K}$ is closed in $\rightset X$
For brevity write: 
:$\ds \SS_1 = \leftset {K \supseteq A: K}$ is closed in $\rightset X$
and:
:$\ds \SS_2 = \leftset {K \supseteq \lambda A: K}$ is closed in $\rightset X$
so that: 
:$\ds A^- = \bigcap_{K \mathop \in \SS_1} K$
and:
:$\ds \paren {\lambda A}^- = \bigcap_{K' \mathop \in \SS_2} K'$
From Dilation of Intersection of Subsets of Vector Space, it now suffices to show that: 
:$\SS_2 = \set {\lambda K : K \in \SS_1}$.
Let $K \in \SS_1$. 
Then $K$ is closed and $A \subseteq K$. 
From Dilation of Closed Set in Topological Vector Space is Closed Set, $\lambda K$ is closed.
Since $\lambda A \subseteq \lambda K$, we have: 
:$\set {\lambda K: K \in \SS_1} \subseteq \SS_2$
Now let $K' \in \SS_2$.
Then $K'$ is closed and $\lambda A \subseteq K'$. 
Let $K = \lambda^{-1} K'$. 
From Dilation of Closed Set in Topological Vector Space is Closed Set, $K$ is closed.
We have $A \subseteq K$, so $K \in \SS_1$. 
Since we have $K' = \lambda K$ for $K \in \SS_1$, we have: 
:$K' \in \set {\lambda K: K \in \SS_1}$
giving:
:$\SS_2 \subseteq \set {\lambda K: K \in \SS_1}$
So:
:$\SS_2 = \set {\lambda K: K \in \SS_1}$
by the definition of set equality.
So: 
Category:Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
Dilation of Closure of Set in Topological Vector Space is Closure of Dilation/Proof 2

== Theorem ==
== Proof ==
Let $x \in \lambda A^-$.
Then $\lambda^{-1} x \in A^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $A$ converging to $\lambda^{-1} x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda y_\mu}_{\mu \mathop \in \Lambda}$ converges to $x$.
Since $\lambda y_\mu \in \lambda A$ for each $\mu \in \Lambda$, we have that $x \in \paren {\lambda A}^-$ from Point in Set Closure iff Limit of Net.
So $\lambda A^- \subseteq \paren {\lambda A}^-$. 
Now let $x \in \paren {\lambda A}^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $\lambda A$ converging to $x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda^{-1} y_\mu}_{\mu \mathop \in \Lambda}$ converges to $\lambda^{-1} x$.
Since $\lambda^{-1} y_\mu \in A$ for each $\mu \in \Lambda$, we have that $\lambda^{-1} x \in A^-$ from Point in Set Closure iff Limit of Net..
So $x \in \lambda A^-$.
Hence we have $\paren {\lambda A}^- \subseteq \lambda A^-$.
Hence we conclude that $\lambda A^- = \paren {\lambda A}^-$.
Category:Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
Dilation of Compact Set in Topological Vector Space is Compact

== Theorem ==
Let $k$ be a topological field.
Let $X$ be a topological vector space over $X$. 
Let $K$ be a compact subset of $X$. 
Let $t \in k \setminus \set {0_k}$. 
Then $t K$ is compact.
== Proof 1 ==
== Proof 2 ==
Category:Compact Spaces
Category:Topological Vector Spaces
Category:Dilation of Subsets of Vector Spaces
Category:Dilation of Compact Set in Topological Vector Space is Compact
Dilation of Compact Set in Topological Vector Space is Compact/Proof 1

== Theorem ==
== Proof ==
Let $\family {U_\alpha : \alpha \in I}$ be open sets such that: 
:$\ds t K \subseteq \bigcup_{\alpha \mathop \in I} U_\alpha$
From Dilation of Union of Subsets of Vector Space, we have: 
:$\ds K \subseteq \bigcup_{\alpha \mathop \in I} \paren {t^{-1} U_\alpha}$ 
From Dilation of Open Set in Topological Vector Space is Open, we have that $t^{-1} U_\alpha$ is open.
Since $K$ is compact, there exists $t^{-1} U_1, t^{-1} U_2, \ldots, t^{-1} U_n$ such that: 
:$\ds K \subseteq \bigcup_{i \mathop = 1}^n \paren {t^{-1} U_i}$ 
Then, using Dilation of Union of Subsets of Vector Space again we have: 
:$\ds t K \subseteq \bigcup_{i \mathop = 1}^n U_i$ 
So every open cover of $t K$ has a finite subcover.
So $t K$ is compact.
Category:Dilation of Compact Set in Topological Vector Space is Compact
Dilation of Compact Set in Topological Vector Space is Compact/Proof 2

== Theorem ==
== Proof ==
From Dilation Mapping on Topological Vector Space is Continuous, the mapping $c_t : X \to X$ defined by:
:$\map {c_t} x = t x$
for each $x \in X$ is continuous.
From Continuous Image of Compact Space is Compact:
:$\map {c_t} K = t K$ is compact.
Category:Dilation of Compact Set in Topological Vector Space is Compact
Dilation of Complement of Set in Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $\lambda \in K$.
Then:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$
== Proof ==
It is immediate from the definition of a dilation that if $x \in F$ we have $\lambda x \in \lambda F$. 
Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.
That is, $\lambda \paren {x - y} = 0$. 
Since $\lambda \ne 0$, it follows that $x = y$, and so $x \in F$.
So we have $x \in F$ $\lambda x \in \lambda F$.
So for $x \in X$ we have $x \not \in F$ $\lambda x \not \in \lambda F$. 
That is, $x \in X \setminus F$ $\lambda x \in X \setminus \paren {\lambda F}$.
It follows that $\lambda x \in \lambda \paren {X \setminus F}$ $\lambda x \in X \setminus \paren {\lambda F}$.
We can therefore conclude:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$
Category:Dilation of Subsets of Vector Spaces
Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $C$ be a convex set with ${\mathbf 0}_X \in C$.
Let $t \in \closedint 0 1$.
Then:
:$t C \subseteq C$
== Proof ==
Let $x \in t C$.
Then we have $x \in t C + \paren {1 - t} {\mathbf 0}_X$.
Since ${\mathbf 0}_X \in C$, we have $x \in t C + \paren {1 - t} C$.
By definition 2 of a convex set, we have $t C + \paren {1 - t} C \subseteq C$.
So we have $x \in C$.
So $x \in t C$ implies that $x \in C$.
So $t C \subseteq C$.
Category:Convex Sets (Vector Spaces)
Dilation of Convex Set in Vector Space is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a vector space over $\Bbb F$. 
Let $C \subseteq X$ be a convex subset of $X$.
Let $\alpha \in \Bbb F$. 
Then $\alpha C$ is convex.
== Proof ==
Consider first the case $\alpha = 0$.
We then have $\alpha C = \set { {\mathbf 0}_X}$.
This is convex by Singleton is Convex Set.
Now consider the case $\alpha \ne 0$. 
Let $u, v \in \alpha C$ and $t \in \closedint 0 1$. 
Then there exists $x, y \in C$ such that $u = \alpha x$ and $v = \alpha y$.
Since $C$ is convex, we have: 
:$t x + \paren {1 - t} y \in C$ 
Then, we have: 
:$\alpha t x + \alpha \paren {1 - t} y \in \alpha C$ 
That is: 
:$t u + \paren {1 - t} v = t \paren {\alpha x} + \paren {1 - t} \paren {\alpha y} \in \alpha C$
Since $u, v \in \alpha C$ and $t \in \closedint 0 1$ were arbitrary, $\alpha C$ is convex.
Category:Convex Sets (Vector Spaces)
Category:Dilation of Subsets of Vector Spaces
Dilation of Interior of Set in Topological Vector Space is Interior of Dilation

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$.
Let $\lambda \in K \setminus \set 0$.
Then we have: 
:$\lambda A^\circ = \paren {\lambda A}^\circ$
where $A^\circ$ denotes the intrerior of $A$. 
== Proof ==
From the definition of intrerior we have:
:$\ds A^\circ = \bigcup \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$
For brevity write: 
:$\ds \SS_1 = \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
:$\ds \SS_2 = \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$
so that: 
:$\ds A^\circ = \bigcup_{U \mathop \in \SS_1} U$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup_{U \mathop \in \SS_2} U$
From Dilation of Union of Subsets of Vector Space, it now suffices to show that: 
:$\SS_2 = \set {\lambda U : U \in \SS_1}$
Let $U \in \SS_1$.
Then $U$ is open and $U \subseteq A$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda U$ is open.
Since $\lambda U \subseteq \lambda A$, we have: 
:$\set {\lambda U: U \in \SS_1} \subseteq \SS_2$
Now let $U' \in \SS_2$.
Then $U'$ is open and $U' \subseteq \lambda A$.
Let $U = \lambda^{-1} U'$.
From Dilation of Open Set in Topological Vector Space is Open, $U$ is open.
We have $U \subseteq A$, so $U \in \SS_1$. 
Since we have $U' = \lambda U$ for $U \in \SS_1$, we have: 
:$U' \in \set {\lambda U: U \in \SS_1}$
giving:
:$\SS_2 \subseteq \set {\lambda U : U \in \SS_1}$
So:
:$\SS_2 = \set {\lambda U : U \in \SS_1}$
by the definition of set equality.
So: 
Category:Topological Vector Spaces
Category:Set Interiors
Category:Dilation of Subsets of Vector Spaces
Dilation of Intersection of Subsets of Vector Space

== Theorem ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.
== Proof ==
First, if $\lambda = 0_K$ then we have:
:$\lambda E_\alpha = \set { {\mathbf 0}_X}$
and:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \set { {\mathbf 0}_X}$
so that:
:$\ds \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha} = \set { {\mathbf 0}_X} = \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
Now take $\lambda \ne 0_K$.
Let $v \in X$.
Then, we have:
:$\ds v \in \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
:
:$\ds \lambda^{-1} v \in \bigcap_{\alpha \mathop \in I} E_\alpha$
:
:$\lambda^{-1} v \in E_\alpha$ for each $\alpha \in I$
:
:$v \in \lambda E_\alpha$ for each $\alpha \in I$
:
:$\ds v \in \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
So, we have:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
in the case $\lambda \ne 0_K$ as well.
Category:Dilation of Subsets of Vector Spaces
Category:Set Intersection
Dilation of Open Set in Normed Vector Space is Open

== Theorem ==
Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $U \subseteq X$ be an open set.
Let $\lambda \in \Bbb F \setminus \set 0$. 
Then:
:$\lambda U$ is open.
== Proof ==
Let:
:$v \in \lambda U$
Then:
:$\dfrac v \lambda \in U$
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:
:$\ds \norm {\frac v \lambda - \frac {v'} \lambda} < \epsilon$
we have $v'/\lambda \in U$. 
That is, $v' \in \lambda U$. 
So, whenever: 
:$\ds \norm {v - v'} < \epsilon \cmod \lambda$
we have $v' \in \lambda U$.
Since $v$ was arbitrary: 
:$\lambda U$ is open.
Category:Open Sets (Normed Vector Spaces)
Category:Dilation of Subsets of Vector Spaces
Dilation of Open Set in Topological Vector Space is Open

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $\lambda \in K \setminus \set {0_K}$. 
Let $U$ be a open set in $X$.
Then $\lambda U$, the dilation of $U$ by $\lambda$ is open.
== Proof ==
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.
Hence the image $c_\lambda \sqbrk U$ is open.
That is, $\lambda U$ is open.
Category:Topological Vector Spaces
Dilation of Subset of Vector Space Distributes over Sum

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
=== Finite Case ===
=== General Case ===
Category:Dilation of Subsets of Vector Spaces
Category:Linear Combination of Subsets of Vector Spaces
Category:Dilation of Subset of Vector Space Distributes over Sum
Dilation of Subset of Vector Space Distributes over Sum/Finite Case

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $A_1, \ldots, A_n \subseteq X$ and $\lambda \in \GF$. 
Then:
:$\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$
where:
:$\ds \lambda \paren \ldots$ denotes dilation by $\lambda$
:$\ds \sum_{j \mathop = 1}^n A_j$ denotes the linear combination of subsets of a vector space. 
== Proof ==
Let $x \in X$.
We have:
:$\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$
 there exists $x_j \in A_j$ for each $j \in \set {1, 2, \ldots, n}$ such that:
:$\ds x = \lambda \sum_{j \mathop = 1}^n x_j$
This is equivalent to:
:$\ds x = \sum_{j \mathop = 1}^n \lambda x_j \in \sum_{j = 1}^n \paren {\lambda A_j}$ for some $x_j \in A_j$. 
Hence we have:
:$\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$
:
:$\ds x \in \sum_{j = 1}^n \paren {\lambda A_j}$
Hence we obtain:
:$\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$
Category:Dilation of Subset of Vector Space Distributes over Sum
Dilation of Subset of Vector Space Distributes over Sum/General Case

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $\family {E_\alpha}_{\alpha \mathop \in A}$ be an $A$-indexed family of sets. 
Let $\lambda \in K$.
Then:
:$\ds \lambda \sum_{\alpha \mathop \in A} E_\alpha = \sum_{\alpha \mathop \in A} \paren {\lambda E_\alpha}$
== Proof ==
Let $x \in X$.
We have:
:$\ds x \in \lambda \sum_{\alpha \mathop \in A} E_\alpha$
 there exists:
:a finite subset $F \subseteq E_\alpha$
:$x_\alpha \in F$ for each $\alpha \in F$
such that:
:$\ds x = \lambda \sum_{\alpha \in F} x_\alpha$
This is equivalent to:
:$\ds x = \sum_{\alpha \in F} \lambda x_\alpha$
for a finite subset $F \subseteq E_\alpha$, with $x_\alpha \in F$ for each $\alpha \in F$. 
Hence we obtain:
:$\ds x \in \lambda \sum_{\alpha \mathop \in A} E_\alpha$ $x \in \sum_{\alpha \mathop \in A} \paren {\lambda E_\alpha}$
so that:
:$\ds \lambda \sum_{\alpha \mathop \in A} E_\alpha = \sum_{\alpha \mathop \in A} \paren {\lambda E_\alpha}$
Category:Dilation of Subset of Vector Space Distributes over Sum
Dilation of Union of Subsets of Vector Space

== Theorem ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then: 
:$\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.
== Proof ==
We have: 
:$\ds v \in \lambda \bigcup_{\alpha \mathop \in I} E_\alpha$
:
:$v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to: 
:there exists some $\alpha \in I$ such that $v = \lambda x$ for some $x \in E_\alpha$.
This is equivalent to: 
:there exists some $\alpha \in I$ such that $v \in \lambda E_\alpha$.
This is finally equivalent to: 
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
So by the definition of set equality, we have: 
:$\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
Category:Dilation of Subsets of Vector Spaces
Category:Set Union
Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$. 
Let $\sequence {\delta_n}_{n \mathop \in \N}$ be a strictly decreasing real sequence with $\delta_n \to 0$.
Let $V$ be a von Neumann-bounded open neighborhood of ${\mathbf 0}_X$.
Then: 
:$\BB = \set {\delta_n V : n \in \N}$ is a local basis for ${\mathbf 0}_X$.
== Proof ==
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
We show that there exists $N \in \N$ such that: 
:$\delta_N V \subseteq U$ 
Since $V$ is von Neumann-bounded, there exists $s > 0$ such that: 
:$V \subseteq t U$ for $t > s$.
Since $\delta_n \to 0$ and $\sequence {\delta_n}_{n \mathop \in \N}$ is a strictly decreasing real sequence, there exists $N \in \N$ with: 
:$\ds 0 < \delta_N < \frac 1 s$
so that: 
:$\ds \frac 1 {\delta_N} > s$ 
giving:
:$\ds V \subseteq \frac 1 {\delta_N} U$
so that: 
:$\delta_N V \subseteq U$
Since $U$ was arbitrary, we have that $\BB$ is a local basis for ${\mathbf 0}_X$.
== Sources ==
* : $1.15$: Theorem
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Category:Dilation of Subsets of Vector Spaces
Category:Local Bases