avewright/proofwiki-math huggingface dataset:
Theorem: Countable Union of Meager Sets is Meager

Let $\struct {X, \tau}$ be a topological space.
Let $\family {A_n}_{n \in \N}$ be a countable set of meager subsets of $X$.
Let:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$ 
Then $A$ is meager in $X$.

Proof:
For each $n \in \N$, $A_n$ is meager and hence there exists a countable set $\family {A_{n, m} }_{m \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A_n = \bigcup_{m \mathop = 1}^\infty A_{n, m}$
Hence:
:$\ds A = \bigcup_{n \mathop = 1}^\infty \bigcup_{m \mathop = 1}^\infty A_{n, m} = \bigcup_{\tuple {n, m} \in \N^2} A_{n, m}$
From Cartesian Product of Countable Sets is Countable, $A$ is therefore the countable union of nowhere dense sets in $X$.
{{mistake|Should this not invoke Countable Union of Countable Sets is Countable I see no cartesian products?}}
So $A$ is meager in $X$.
{{qed}}
{{ACC|Countable Union of Countable Sets is Countable}}
Theorem: Join Semilattice Ideal iff Ordered Set Ideal

Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
Then:
:$I$ is a join semilattice ideal of $\struct {S, \vee, \preceq}$ {{iff}} $I$ is an ordered set ideal of $\struct {S, \preceq}$.

Proof:
=== Necessary Condition ===
Let $I$ be a join semilattice ideal of $\struct {S, \vee, \preceq}$.
To show that $I$ is an ordered set ideal of $\struct {S, \preceq}$ it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a directed subset of $S$:
        | q = \forall x, y \in I: \exists z \in I
        | m = x \preceq z \text{ and } y \preceq z
}}
{{end-axiom}}
Let $x, y \in I$.
Let $z = x \vee y$.
By definition of join semilattice ideal, $I$ is a subsemilattice, so:
:$z \in I$
By definition of join:
:$x \preceq z \text{ and } y \preceq z$
The result follows.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Lattice Ideal

Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $I \subseteq S$ be a non-empty subset of $S$.
{{TFAE|def = Lattice Ideal}}

Proof:
=== Definition 1 implies Definition 2 ===
Let $I$ satisify the lattice ideal axioms.
To show that $I$ is a join semilattice ideal it is sufficient to show:
{{begin-axiom}}
{{axiom | lc= $I$ is a lower section of $S$:
        | q = \forall x \in F: \forall y \in S
        | m = y \preceq x \implies y \in I
}}
{{end-axiom}}
Let $x \in I, y \in S : y \preceq x$.
By the lattice ideal axioms, $F$ is a sublattice of $\struct {S, \vee, \wedge, \preceq}$, so:
:$x \wedge y \in I$
From Preceding iff Meet equals Less Operand:
:$y = x \wedge y$
Hence:
:$y \in I$
The result follows.
{{qed|lemma}}
Theorem: Finite Subset of Topological Vector Space is von Neumann-Bounded

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$. 
Let $F$ be a finite subset of $X$. 
Then $F$ is von Neumann-bounded.

Proof:
Let $F = \set {v_1, \ldots, v_m}$.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Multiple of Vector in Topological Vector Space Converges, we have: 
:$\ds \frac {v_i} n \to {\mathbf 0}_X$ as $n \to \infty$
for each $1 \le i \le n$.
Hence for each $1 \le i \le n$, there exists $N_i \in \N$ such that:
:$\ds \frac {v_i} n \in U$ for $n \ge N_i$
Let:
:$N = \max \set {N_1, \ldots, N_m}$
Then for $n \ge N$ we have:
:$v_i \in n U$ for each $1 \le i \le n$.
That is:
:$F \subseteq n U$ for each $1 \le i \le n$.
Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $F$ is von Neumann-bounded.
{{qed}}
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
ncvomkilvwx6p0p5q9n4049viatlunv
Theorem: Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$. 
Then: 
:$T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$.

Proof:
The result is immediate when $\lambda = 0_K$, since $T^{-1} \sqbrk {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$.
Now take $\lambda \ne 0_K$.
Let $x \in X$.
We have:
:$x \in T^{-1} \sqbrk {\lambda E}$
{{iff}}:
:$T x \in \lambda E$
From linearity, this is the case {{iff}}:
:$\map T {\lambda^{-1} x} \in E$
This is equivalent to:
:$\lambda^{-1} x \in T^{-1} \sqbrk E$
Which in turn is equivalent to:
:$x \in \lambda T^{-1} \sqbrk E$
So we obtain:
:$T^{-1} \sqbrk {\lambda E} = \lambda T^{-1} \sqbrk E$
{{qed}}
Category:Dilations of Subsets of Vector Spaces
ibnl3i5zelmib07m2mgma2ojgdiax29
Theorem: Homeomorphic Image of Meager Set is Meager

Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A \subseteq X$ be meager in $X$.
Then $f \sqbrk A$ is meager in $Y$.

Proof:
Since $A$ is meager in $X$, there exists a countable set $\family {A_n}_{n \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
So, we have:
:$\ds f \sqbrk A = f \sqbrk {\bigcup_{n \mathop = 1}^\infty A_n}$
So, from Image of Union under Mapping:
:$\ds f \sqbrk A = \bigcup_{n \mathop = 1}^\infty f \sqbrk {A_n}$ 
From Homeomorphic Image of Nowhere Dense Set is Nowhere Dense, $f \sqbrk {A_n}$ is nowhere dense in $Y$ for each $n \in \N$. 
Hence $f \sqbrk A$ is the countable union of nowhere dense sets in $Y$.
So $f \sqbrk A$ is meager in $Y$. 
{{qed}}
Category:Meager Spaces
j6b280kdd8ywyz32kvwoaijpc0sadv8
Theorem: Banach-Steinhaus Theorem/F-Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau_X}$ be an $F$-Space over $\GF$.
Let $\struct {Y, \tau_Y}$ be a topological vector space over $\GF$.
Let $\Gamma$ be a set of continuous linear transformations $X \to Y$ such that for all $x \in X$:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in $Y$.
Then $\Gamma$ is equicontinuous.

Proof:
Let $B$ be the set of $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in $Y$.
By hypothesis, $B = X$. 
Let $d$ be a metric inducing $\tau_X$ such that $\struct {X, d}$ is a complete metric space. 
From the Baire Category Theorem, $\struct {X, d}$ is a Baire space.
From Baire Space is Non-Meager, $\struct {X, d}$ is not meager.
Since $B$ is then not meager, we can apply the Banach-Steinhaus Theorem to obtain that $\Gamma$ is equicontinuous.
{{qed}}
Theorem: Norm Topology Induced by Metric Induced by Norm

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $d$ be the metric induced on $X$ by $\norm {\, \cdot \,}$.
Then the open sets of $\struct {X, d}$ as a metric space are precisely the open sets of $\struct {X, \norm {\, \cdot \,} }$ as a normed vector space.

Proof:
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the open ball in $\struct {X, d}$ with radius $\epsilon$ and center $x$. 
For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^{\norm {\, \cdot \,} } } x$ be the open ball in $\struct {X, \norm {\, \cdot \,} }$ with radius $\epsilon$ and center $x$. 
{{Recall|Open Set of Metric Space}}
{{:Definition:Open Set of Metric Space}}
{{Recall|Open Set in Normed Vector Space}}
{{:Definition:Open Set in Normed Vector Space}}
It is therefore sufficient to show that $\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$.
Let $y \in X$.
We have $y \in \map {B_\epsilon^d} x$ {{iff}}:
:$\map d {x, y} < \epsilon$
This is equivalent to:
:$\norm {x - y} < \epsilon$
This is then equivalent to:
:$y \in \map {B_\epsilon^{\norm {\, \cdot \,} } } x$
Hence we have: 
:$\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$
{{qed}}
Category:Normed Vector Spaces
Category:Metric Spaces
cxccff39jj2q2agrz0xx639zmog9vmc
Theorem: Banach Space is F-Space

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$. 
From Normed Vector Space is Hausdorff Topological Vector Space, we may consider $\struct {X, \norm {\, \cdot \,} }$ as a topological vector space.
With this identification, $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.

Proof:
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
From Norm Topology Induced by Metric Induced by Norm, $d$ induces the topology on $\struct {X, \norm {\, \cdot \,} }$.
Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\struct {X, d}$ is a complete metric space. 
From Metric Induced by Norm is Invariant Metric, $d$ is invariant.
Hence $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.
{{qed}}
Category:F-Spaces
Category:Banach Spaces
gdxkib1zxm38y8834b0siwjx2raqqwq
Theorem: Proper Closed Linear Subspace of Topological Vector Space is Meager

Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$. 
Let $D$ be a proper closed linear subspace of $X$.
Then $D$ is meager.

Proof:
From Set is Closed iff Equals Topological Closure, we have $D^- = D$. 
From Proper Linear Subspace of Topological Vector Space has Empty Interior, we then have that $\paren {D^-}^\circ = D^\circ = \O$. 
Hence $D$ is nowhere dense, and in particular meager.
{{qed}}
Category:Meager Spaces
Category:Topological Vector Spaces
fm2khrqo0p3jkd9urxw7mfw6kaww551
Theorem: Proper Linear Subspace of Topological Vector Space has Empty Interior

Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper linear subspace of $X$. 
Then:
:$L^\circ = \O$
where $L^\circ$ denotes the interior of $L$.

Proof:
{{AimForCont}} that $L^\circ \ne \O$.
Let $x \in L^\circ$. 
Then there exists an open neighborhood $U$ of $x$ such that $U \subseteq L$. 
From Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of ${\mathbf 0}_X$ with $U - x \subseteq L - x$.
Since $L$ is a linear subspace of $X$, we have $L - x = L$.
Let $V = U - x$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:
:$\ds X = \bigcup_{n \mathop = 1}^\infty n V$
On the other hand, since $L$ is a linear subspace of $X$ and $V \subseteq L$, we also have:
:$\ds \bigcup_{n \mathop = 1}^\infty n V \subseteq L$
and so $L = X$.
This contradicts that $L$ is a proper linear subspace.
Hence $L^\circ = \O$.
{{qed}}
Category:Topological Vector Spaces
bs4msx79r5f3dd08wft2n70o8snt7y7
Theorem: Non-Meager Linear Subspace of Topological Vector Space is Everywhere Dense

Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper non-meager linear subspace of $X$. 
Then $L$ is everywhere dense.

Proof:
{{AimForCont}} that $L$ is not everywhere dense.
Then $L^- \ne X$.
From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $L^-$ is then a proper closed linear subspace of $X$.
From Proper Closed Linear Subspace of Topological Vector Space is Meager, it follows that $L^-$ is meager. 
From Subset of Meager Set is Meager Set, $L$ is meager.
This is contrary to our assumption that $L$ is non-meager.
So $L$ is everywhere dense.
{{qed}}
Category:Non-Meager Spaces
Category:Everywhere Dense
Category:Topological Vector Spaces
0m89hlxo2tnbnzp3yakimg4wvg4kicc
Theorem: Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence in $X$.
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a Cauchy sequence in $X$.

Proof:
First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X$ as $n \to \infty$.
From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {\lambda x_n}_{n \in \N}$ is therefore Cauchy.
Now take $\lambda \ne 0$. 
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda^{-1} U$ is an open neighborhood of ${\mathbf 0}_X$.
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that:
:$x_n - x_m \in \lambda^{-1} U$ for $n, m \ge N$.
We therefore have:
:$\lambda x_n - \lambda x_m \in U$ for $n, m \ge N$.
Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $\sequence {\lambda x_n}_{n \in \N}$ is Cauchy.
{{qed}}
Category:Cauchy Sequences in Topological Vector Spaces
a57eyx0lr1wxp72yxb35icmqvm3uepk
Theorem: Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be Cauchy sequences in $X$.
Then $\sequence {x_n + y_n}_{n \in \N}$ is a Cauchy sequence in $X$.

Proof:
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary 1, there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:
:$V + V \subseteq U$
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N_1 \in N$ such that:
:$x_n - x_m \in V$ for $n, m \ge N_1$.
Since $\sequence {y_n}_{n \in \N}$ is Cauchy, there exists $N_2 \in \N$ such that:
:$y_n - y_m \in V$ for $n, m \ge N_2$.
Let $N = \max \set {N_1, N_2}$. 
Then, for $n, m \ge N$, we have: 
{{begin-eqn}}
{{eqn	| l = \paren {x_n + y_n} - \paren {x_m + y_m}
		| r = \paren {x_n - x_m} + \paren {y_n - y_m}
}}
{{eqn	| o = \in
		| r = V + V
}}
{{eqn	| o = \subseteq
		| r = U
}}
{{end-eqn}}
Hence $\sequence {x_n + y_n}_{n \in \N}$ is a Cauchy sequence in $X$.
{{qed}}
Category:Cauchy Sequences in Topological Vector Spaces
53h2p4bq3pmw4r7ss2de2h07e3t3y1u
Theorem: Linear Combination of Cauchy Sequences in Topological Vector Space is Cauchy Sequence

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be Cauchy sequences in $X$.
Let $\lambda, \mu \in \GF$.
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is a Cauchy sequence in $X$.

Proof:
From Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n}_{n \in \N}$ and $\sequence {\mu y_n}_{n \in \N}$ are Cauchy in $X$.
From Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is Cauchy.
{{qed}}
Category:Cauchy Sequences in Topological Vector Spaces
69u9xrxqd03h0yud4t60qi5tt19ej40
Theorem: Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x \in X$.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$ with:
:$x_n \to x$
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a convergent sequence in $X$ with:
:$\lambda x_n \to \lambda x$.

Proof:
First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X = \lambda x$ as $n \to \infty$.
So $\sequence {\lambda x_n}_{n \in \N}$ converges for $\lambda = 0$. 
Now take $\lambda \ne 0_K$.
Let $U$ be an open neighborhood of $\lambda x$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda^{-1} U$ is an open neighborhood of $x$.  
Since $\sequence {x_n}_{n \in \N}$ converges, there exists $N \in \N$ such that:
:$x_n \in \lambda^{-1} U$ for $n \ge N$.
Then, we have:
:$\lambda x_n \in U$ for $n \ge N$. 
Since $U$ was an arbitrary open neighborhood of $\lambda x$, we have that  $\sequence {\lambda x_n}_{n \in \N}$ converges to $\lambda x$ as required.
{{qed}}
Category:Topological Vector Spaces
du8p3e6s0txa30u7jmz95erqm718fqg
Theorem: Sum of Convergent Sequences in Topological Vector Space is Convergent

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {x_n + y_n}_{n \in \N}$ converges with:
:$x_n + y_n \to x + y$

Proof:
Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subseteq W$
Since $\sequence {x_n}_{n \in \N}$ converges to $x$, there exists $N_1 \in \N$ such that:
:$x_n \in U$ for $n \ge N_1$.
Since $\sequence {y_n}_{n \in \N}$ converges to $y$, there exists $N_2 \in \N$ such that:
:$y_n \in V$ for $n \ge N_2$.
Let $N = \max \set {N_1, N_2}$. 
Then for $n \ge N$ we have: 
:$x_n + y_n \in U + V \subseteq W$
Since $W$ was an arbitrary open neighborhood of $x + y$, we have that $\sequence {x_n + y_n}$ converges to $x + y$.
{{qed}}
Category:Topological Vector Spaces
3xtlgpr6he8iop4fek4iz2ekit73pnj
Theorem: Linear Combination of Convergent Sequences in Topological Vector Space is Convergent

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$ and $\lambda, \mu \in K$. 
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ converges with:
:$\lambda x_n + \mu y_n \to \lambda x + \mu y$

Proof:
From Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent, we have that:
:$\sequence {\lambda x_n}_{n \in \N}$ converges to $\lambda x$
and:
:$\sequence {\mu y_n}_{n \in \N}$ converges to $\mu y$.
From Sum of Convergent Sequences in Topological Vector Space is Convergent, $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ converges to $\lambda x + \mu y$.
{{qed}}
Category:Topological Vector Spaces
8d5ebme920ohz6z0vlthx56ou03slle
Theorem: Constant Sequence in Topological Space Converges

Let $X$ be a topological space. 
Let $x \in X$. 
Define a sequence $\sequence {x_n}_{n \in \N}$ by:
:$x_n = x$ for each $n \in \N$. 
Then $\sequence {x_n}_{n \in \N}$ converges to $x$.

Proof:
Let $U$ be an open neighborhood of $x$.
Then we have $x_n = x \in U$ for all $n \in \N$. 
Hence $\sequence {x_n}_{n \in \N}$ converges to $x$.
{{qed}}
Category:Convergence
mww034dbrt84w84623o1xuwp9ay3x6p
Theorem: Sum of Convergent Nets in Topological Vector Space is Convergent

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x, y \in X$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ and $\family {y_\lambda}_{\lambda \in \Lambda}$ be nets converging to $x$ and $y$ respectively.
Then the net $\family {x_\lambda + y_\lambda}_{\lambda \in \Lambda}$ converges to $x + y$.

Proof:
For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subseteq W$
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_1 \in \Lambda$ such that:
:$x_\lambda \in U$ for $\lambda \succeq \lambda_1$.
Since $\family {y_\lambda}_{\lambda \in \Lambda}$ converges to $y$, there exists $\lambda_2 \in \Lambda$ such that:
:$y_\lambda \in V$ for $\lambda \succeq \lambda_2$.
Since $\struct {\Lambda, \preceq}$ is directed, there exists $\lambda_\ast \in \Lambda$ such that: 
:$\lambda_\ast \succeq \lambda_1$ and $\lambda_\ast \succeq \lambda_2$.
Since $\preceq$ is transitive, if $\lambda \in \Lambda$ has $\lambda \succeq \lambda_\ast$, then $\lambda \succeq \lambda_1$ and $\lambda \succeq \lambda_2$.
Hence, for $\lambda \succeq \lambda_\ast$ we have $x_\lambda \in U$ and $y_\lambda \in V$.
Hence:
:$x_\lambda + y_\lambda \in U + V \subseteq W$ for $\lambda \succeq \lambda_\ast$. 
Since $W$ was an arbitrary open neighborhood of $x + y$, the net $\family {x_\lambda + y_\lambda}_{\lambda \in \Lambda}$ converges to $x + y$. 
{{qed}}
Category:Nets (Set Theory)
Category:Topological Vector Spaces
3iq4njanm3y6sp4wmajum6f72o0txz7
Theorem: Scalar Multiple of Convergent Net in Topological Vector Space is Convergent

Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$ and $\mu \in K$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net converging to $x$.
Then $\family {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$.

Proof:
For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
First consider the case $\mu = 0_K$.
Then $\mu x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Net is Convergent, we have that $\mu x_\lambda$ converges to ${\mathbf 0}_X$. 
So $\sequence {\lambda x_n}_{n \in \N}$  converges to $\mu x$ for $\mu = 0_K$.
Now take $\mu \ne 0_K$.
Let $U$ be an open neighborhood of $\mu x$.
From Dilation of Open Set in Topological Vector Space is Open, $\mu^{-1} U$ is an open neighborhood of $x$.  
Since $\sequence {x_n}_{n \in \N}$ converges, there exists $\lambda_0 \in \Lambda$ such that:
:$x_n \in \mu^{-1} U$ for $\lambda \succeq \lambda_0$.
Then, we have:
:$\mu x_n \in U$ for $\lambda \succeq \lambda_0$. 
Since $U$ was an arbitrary open neighborhood of $\mu x$, we have that  $\sequence {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$ as required.
{{qed}}
Category:Nets (Set Theory)
Category:Topological Vector Spaces
mygw8rit4x1pokvnzwakejnb66dljgk
Theorem: Characterization of Convex Absorbing Set in Vector Space

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $C$ be a convex set. 
Then $C$ is absorbing {{iff}}:
:for each $x \in X$ there exists $t \in \R_{> 0}$ such that $x \in t C$.

Proof:
=== Necessary Condition ===
Suppose that $C$ is absorbing.
Then for each $x \in X$ there exists $s \in \R_{> 0}$ such that $x \in t C$ for $s \in \C$ with $\cmod s \ge t$.
In particular, $x \in t C$. 
{{qed|lemma}}
Theorem: Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1

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 $t C \subseteq C$.
{{qed}}
Category:Convex Sets (Vector Spaces)
57pqkcm1jq8fleoq4j1i1cwzbzq8ljq
Theorem: Superset of Absorbing Set is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A$ be an absorbing set.
Let $B \supseteq A$. 
Then $B$ is absorbing.

Proof:
Let $x \in X$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in t A$ for $\cmod \alpha \ge t$.
Since $A \subseteq B$, we obtain:
:$x \in t B$ for $\cmod \alpha \ge t$.
{{qed}}
Category:Absorbing Sets
42aktvnfybdn2rnwsciepjkibi8qpw5
Theorem: Image of Absorbing Set under Surjective Linear Transformation is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Let $T : X \to Y$ be a surjective linear transformation. 
Then $T \sqbrk A$ is an absorbing set.

Proof:
Let $y \in Y$.
Since $T$ is surjective, there exists $x \in X$ such that $y = T x$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
Then:
:$y = T x \in T \sqbrk {\alpha A}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$. 
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:
:$y \in \alpha T \sqbrk A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$. 
So $T \sqbrk A$ is absorbing.
{{qed}}
Category:Absorbing Sets
1234o6zjht0a9gsgq4akxlu2vxayg5c
Theorem: Closure of Absorbing Set is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Then $A^-$ is absorbing.

Proof:
From the definition of closure, we have $A \subseteq A^-$.
From Superset of Absorbing Set is Absorbing, $A^-$ is absorbing.
{{qed}}
Category:Absorbing Sets
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