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Let $X$ and $Y$ be Banach space. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. Also, $K(X,Y)$ is the set of all compact operators from $X$ to $Y$.

Why is $K(X ,Y)$ is a closed subspace of $B(X, Y)$?

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Let $T_i \in K$ and $T_i \to T$ in norm topology. Let $\epsilon >0$. Let $x_j$ be a bounded sequence in $X$. As $T_k$ is compact for each $k$, using a diagonal sequence arguement, there is a subsequence of $\{x_n\}$ which we still call the same sequence so that $\{T_k x_n\}$ is convergent for all $k$. Let $k_0 \in \mathbb N$ such that $||T- T_k|| <\epsilon$. Since $\{T_{k_0} x_n\}$ is convergent, there is $M\in \mathbb N$ such that $||T_{k_0} x_j - T_{k_0} x_l|| < \epsilon $ for all $j, l \geq M$. Then

$$||Tx_j - Tx_l|| \leq \|Tx_j - T_{k_0} x_j\| + \|T_{k_0} x_j - T_{k_0} x_l \| + \|T_{k_0}x_l - T x_l\| \leq (2L+1)\epsilon$$

for all $j, l\geq M$ (where $||x_j|| \leq L$). Thus $\{Tx_j\}_{j=1}^\infty$ is Cauchy. As $Y$ is complete, it is convergent. This shows that $T$ is compact and $K$ is closed.

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    $\begingroup$ Dear John I don't get one thing about your proof. Indeed once you select $\varepsilon > 0$ than by the convergence of the operator you get a different $n\in\mathbb{N}$ and consequentially a different subsequence $\{ x_{k_h} \}_n$ which depends on the current operator. Thus with the previous statement I think you proved that $\forall \varepsilon > 0$ you can always find two elements which image through T is close less than $\varepsilon$. Maybe by taking a sort of diagonal sequence you can conclude, but made this way can it happen that 2 subsequences don't intersect at all? $\endgroup$ – Tommaso Seneci Mar 5 '16 at 10:49
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    $\begingroup$ @TommasoSeneci : This is indeed a good observation. You are right, one need to use a diagonal arguement: there is a subsequence of $x_n$ so that $T_1x_{n_k}$ is convergent. Then with this subsequence, there is a further subsequence so that $T_2(\cdot)$ converges. Then one construct a sequence of subsequence and then take the diagonal subsequence. The final subsequence (rename back to $x_n$) will satisfy that $T_k x_n$ is convergent for all $k$. I will make an edit. $\endgroup$ – user99914 Mar 5 '16 at 11:43

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