# Spectral theory: Operator compact implies existence of convergent subsequence

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.

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.