Spectral theory: Operator compact implies existence of convergent subsequence

Question

So, I'm looking at the proof of the spectral theorem for self-adjoint compact operators in my functional analysis lecture notes (an introductionary class).

We defined a compact operator $T$ as a linear continuous operator between two Hilbert-spaces $X_{1},X_{2}$ hence $T\in L(X_{1},X_{2})$, for which $T(B_{1}(0))$ is totally bounded.

Now in the proof I'd like to know why from the compactness of $T:X\rightarrow X$ it follows that $(T(x_{n}))_{n\in\mathbb{N}}$ has a convergent subsequence in $X$ where $(x_{n})_{n\in\mathbb{N}}$ is a bounded series.

Answer 1

If $(x_n)$ is bounded then $(Tx_n)$ lies in a totally bounded set $E$. The closure of $E$ is a compact set because $X_2$ is a complete metric space. Since a compact metric space is sequentially compact it follows that $(Tx_n)$ has a convergent subsequence.