Theorem: Banach-Steinhaus Theorem/Topological Vector Space

Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be topological vector spaces over $\GF$.
Let $\Gamma$ be a set of continuous linear transformations $X \to Y$.
Let $B$ be the set of all $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$
is von Neumann-bounded in $Y$.
Suppose that $B$ is not meager in $X$.
Then $B = X$ and $\Gamma$ is equicontinuous.

Proof:
Let $W$ be an open neighborhood of ${\mathbf 0}_Y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods, there exists an open neighborhood $V$ of ${\mathbf 0}_Y$ such that:
:$V + V \subseteq W$
From Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary, there exists a further open neighborhood $U$ of ${\mathbf 0}_X$ such that:
:$U^- \subseteq V$
so that:
:$U^- + U^- \subseteq W$
where $U^-$ denotes the closure of $U$ in $Y$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood and Set Closure Preserves Set Inclusion we can take $U$ to be balanced. 
Let:
:$\ds E = \bigcap_{T \in \Gamma} T^{-1} \sqbrk {U^-}$
Since each $T \in \Gamma$ is continuous, and $U^-$ is closed in $Y$ by Topological Closure is Closed we have:
:$T^{-1} \sqbrk {U^-}$ is closed in $Y$ for each $T \in \Gamma$. 
So $E$ is closed in $E$ as the countable intersection of closed sets.
Now let $x \in B$. 
Since $\map \Gamma x$ is von Neumann-bounded in $Y$, there exists $n \in \N$ such that:
:$\map \Gamma x \subseteq n U$
Then we have:
:$T x \in n U$ for each $T \in \Gamma$.
That is:
:$x \in T^{-1} \sqbrk {n U}$
so that:
:$x \in T^{-1} \sqbrk {n U^-}$
for each $T \in \Gamma$. 
Then we have:
{{begin-eqn}}
{{eqn	| l = x
		| o = \in
		| r = \bigcap_{T \in \Gamma} T^{-1} \sqbrk {n U^-}
}}
{{eqn	| r = \bigcap_{T \in \Gamma} n T^{-1} \sqbrk {U^-}
		| c = Preimage of Dilation of Set under Linear Transformation is Dilation of Preimage
}}
{{eqn	| r = n \bigcap_{T \in \Gamma} T^{-1} \sqbrk {U^-}
		| c = Dilation of Intersection of Subsets of Vector Space
}}
{{eqn	| r = n E
}}
{{end-eqn}}
So we may conclude that:
:$\ds B \subseteq \bigcup_{n \mathop = 1}^\infty n E$
We argue that $n E$ is not meager in $X$ for some $n \in \N$.
{{AimForCont}} that $n E$ is meager in $X$ for each $n \in \N$.
Then $\paren {n E} \cap B$ is meager in $X$ for each $n \in \N$ by Subset of Meager Set is Meager Set, while:
:$\ds B = \bigcup_{n \mathop = 1}^\infty \paren {\paren {n E} \cap B}$
From Countable Union of Meager Sets is Meager, we have that $B$ is meager, contrary to our assumption that it is not. 
So we have that $n E$ is meager in $X$ for some $n \in \N$.
From Dilation Mapping on Topological Vector Space is Homeomorphism, the mapping $x \mapsto n^{-1} x$ is a homeomorphism.
From Homeomorphic Image of Meager Set is Meager, we have that $E$ is meager.
Since $E$ is closed, the interior of $E$ is non-empty.
Let $v \in E^\circ$.
Let $V \subseteq E$ be an open neighborhood of $x$.
From Translation of Open Set in Topological Vector Space is Open, $V - x$ is an open neighborhood of ${\mathbf 0}_X$.
By definition of $E$ we have:
:$T v \in U^-$ for all $T \in \Gamma$.
Hence:
{{begin-eqn}}
{{eqn	| l = T \sqbrk {V - x}
		| r = T \sqbrk V - T x
		| c = Image of Translation of Set under Linear Transformation is Translation of Image
}}
{{eqn	| o = \subseteq
		| r = U^- - U^-
}}
{{eqn	| o = \subseteq
		| r = U^- + U^-
		| c = from Closure of Balanced Set in Topological Vector Space is Balanced, $U^-$ is balanced and $-U^- \subseteq U^-$
}}
{{eqn	| o = \subseteq
		| r = W
		| c = by choice of $U$
}}
{{end-eqn}}
for all $T \in \Gamma$. 
Since $W$ was an arbitrary open neighborhood of ${\mathbf 0}_Y$, we have that $\Gamma$ is equicontinuous.
It remains to show that $B = X$.
Let $x \in X$. 
From Finite Subset of Topological Vector Space is von Neumann-Bounded, we have:
:$\set x$ is von Neumann-bounded. 
From Image of von Neumann-Bounded Set under Equicontinuous Family of Linear Transformations is Contained in von Neumann-Bounded Set, there exists a von Neumann-bounded subset $F_x$ of $Y$ such that:
:$T \sqbrk {\set x} \subseteq F_x$
for each $T \in \Gamma$.
That is:
:$T x \in F_x$ for each $T \in \Gamma$
so that:
:$\set {T x : T \in \Gamma} \subseteq F_x$
In other words, from Subset of von Neumann-Bounded Set is von Neumann-Bounded:
:$\set {T x : T \in \Gamma}$ is von Neumann-bounded for each $x \in X$.
So $x \in B$.
Hence we conclude $B = X$.
{{qed}}