Why is the set of compact operators closed in the space of all bounded operators between Banach spaces?

Question

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 $B(X, Y)$?

Answer 1

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.

Answer 2

i know this is a 9 years old question but i want to contribute. Here is another proof of the satement: Let $(f_n)_n$ be a secuence of compact linear maps that converges to $f$, a linear continuous map. Let $A\subseteq X$ be a bounded set, lets say is bounded by $M>0$. I claim that $f[A]$ is totally bounded. Take $r>0$. There exists a natural number $N$ such that $n\geq N \Rightarrow |f_n-f|<r$. This implies that if $a \in A$ and $n\geq N$ then: $$ |f_n a - fa | \leq |f_n - f | |a| < rM. $$ Now, since $f_n[A]$ is totally bounded, there are $a_1,\dots,a_k \in A$ such that $$ \bigcup_{j=1}^k B_r (f_n a_j) \supseteq f_n[A] $$

Let's see that $\bigcup_{j=1}^k B_{r(2M+1)} (fa_j) \supseteq f[A]$.

Take $a \in A$. There is $j_0 \in \{1,\dots,k\}$ such that $f_n a \in B_r(f_n a_{j_0})$. Now, by triangle inequality we get: $$ |fa - f a_{j_0} | \leq |fa - f_n a | + |f_n a - f_n a_{j_0} | + |f_n a_{j_0} - f a_{j_0}| < r(2M +1) $$

Hence $f[A]$ is totally bounded and then $cl (f[A])$ is totally bounded, closed subset of a complete space, that is, compact.