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(Uniform Boundedness Thm) $X : $ Banach space, $Y$ : Normed vector space, $\mathcal A \subset B(X,Y) := \{ f : X \to Y \; | \; f : \text{bounded operator} \} $. If for any $x \in X$, $\sup_{T \in \mathcal A} \| Tx \|_Y < \infty $, then $\sup_{T \in \mathcal A} \| T \| < \infty.$

I am studying this theorem, but in the proof, I could not know that why $F_n := \{ x \in X \; | \; \sup_{T \in \mathcal A} \| Tx \|_Y \leqslant n \} $ is closed for all $n$, and $X = \bigcup_{n=1}^\infty F_n$ .

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$F_n$ is an intersection of closed sets. $X = \cup F_n$ by the assumption that the supremum is finite for all $x$. –  Qiaochu Yuan Jul 19 '12 at 17:33
    
@QiaochuYuan Thank you Qiaochu Yuan, I think that $X = \cup F_n$ is clear, but how can I prove that $F_1 , F_2 , \cdots$ are closed sets respectively? –  Bamily Jul 19 '12 at 18:40
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This is extended version of Qiaochu's comment

1) Note that $$ x\in F_n\Longleftrightarrow \sup_{T\in\mathcal{A}}\Vert Tx\Vert\leq n \Longleftrightarrow \forall T\in\mathcal{A}\quad\Vert Tx\Vert\leq n \Longleftrightarrow\\ \forall T\in\mathcal{A}\quad Tx\in B_Y(0,n) \Longleftrightarrow \forall T\in\mathcal{A}\quad x\in T^{-1}(B_Y(0,n)) \Longleftrightarrow\\ x\in\bigcap_{T\in\mathcal{A}}T^{-1}(B_Y(0,n)) $$ and we conclude $$ F_n=\bigcap_{T\in\mathcal{A}}T^{-1}(B_Y(0,n))\tag{1} $$ The ball $B_Y(0,n)$ is a closed set. Since $T$ is continuous, then preimage $T^{-1}(B_Y(0,n))$ of closed set is closed. Intersection of closed sets is closed, so from $(1)$ we conclude that $F_n$ is closed.

2) Obviously $$ \bigcup\limits_{n\in\mathbb{N}} F_n\subset X\tag{2} $$ Take arbitrary $x\in X$ and consider natural number $N=\lfloor \sup_{T\in\mathcal{A}}\Vert Tx\Vert\rfloor+1$. Then $x\in F_N\subset \bigcup\limits_{n\in\mathbb{N}} F_n$. Since $x\in X$ is arbitrary we see that $$ X\subset \bigcup\limits_{n\in\mathbb{N}} F_n\tag{3} $$ From $(2)$ and $(3)$ the desired equality follows.

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Thank you very much. –  Bamily Jul 19 '12 at 18:52
    
Just to add to this post, another way to see $F_n$ is closed: For any $x \in cl(F_n)$, there is a sequence in $F_n$, say $(x_j)$. This means for every fixed $T \in F_n$, we have that $|Tx_j| \leq n$. Then taking the limits on both sides, we see that $|Tx| \leq n$. This implies $x \in F_n$. –  Tyler Hilton Dec 15 '13 at 9:13
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