# how to prove this property of compact operator？ [duplicate]

$K(X, Y)$ is a closed subspace of $B(X, Y)$: Let $T_{n}, n \in N$, be a sequence of compact operators from one Banach space to the other, and suppose that $T_{n}$, converges to $T$ with respect to the operator norm. Then $T$ is also compact.

Can anybody prove it in details or tell me where I can find the proof? Thanks so much!

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## marked as duplicate by Norbert, 1015, Lost1, T. Bongers, Asaf KaragilaFeb 9 '14 at 6:59

This question was marked as an exact duplicate of an existing question.

See this. – David Mitra Jul 4 '13 at 13:05
In particular, the limit of finite rank operators (not being necessarily finite rank) is compact. And a natural question arises: is every compact operator norm-limit of finite-rank ones? Well it isn't (counterexample by Enflo), and it is something you should know. – sheriff Jul 4 '13 at 13:33

We prove by definition. Let $(x_{n})$ be a bounded sequence. We use the diagonal method to prove that $(Tx_{n})$ has a convergent subsequence.

Since $T_{1}$ is compact, there is a subsequence $(x_{1,n})$ such that $(T_{1}x_{1,n})$ is Cauchy. Similarly, $(x_{1,n})$ has a subsequence $(x_{2,n})$ such that $(T_{2}x_{2,n})$ is Cauchy. Continuing in this way, we see that the diagonal sequence $y_{n} = x_{n,n}$ is a subsequence of $(x_{n})$ such that for every fixed positive integer $m$, the sequence $(T_{m}y_{n})$ is Cauchy. Let $M$ be a bound for the sequence $(x_{n})$. Hence $||y_{n}|| \leq M$ for all $n$. Let $\epsilon > 0$ be given. Since $T_{m} \rightarrow T$, there is a natural number $p$ such that $|| T - T_{p}|| < \frac{\epsilon}{3M}$. Since $(T_{p}y_{n})$ is Cauchy, there is a $N$ such that

$$|| T_{p}y_{j} - T_{p}y_{k} || < \frac{\epsilon}{3M} \quad j,k > N$$

Hence we obtain, for $j,k > N$,

$$||Ty_{j} - Ty_{k}|| \leq ||Ty_{j}- T_{p}y_{j}|| + ||T_{p}y_{j} - T_{p}y_{k} || + ||Ty_{k}- T_{p}y_{k}|| \leq ||T-T_{p}||||y_{j}|| + \frac{\epsilon}{3} + ||T_{p}-T||||y_{k}|| < \epsilon$$.

This shows that $(Ty_{n})$ is Cauchy and hence the proof is complete.

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