Same convergent subsequence for two compact operators?

Question

I was reading a proof on how any linear combination of compact operators is compact.

Let $U,V: X \to Y$ be compact linear operators and let $\alpha,\beta \in \mathbb{C}$. Then each bounded sequence $(x_n)$ in $X$ contains a subsequence $(x_{n(k)})$ such that $(Ax_{n(k)})$ and $(Bx_{n(k)})$ converge. Then the proof says that due to this $(\alpha A + \beta B)x_{n(k)}$ converges.

But what I dont understand is how we have the same indexing for the subsequence in the case of both operators. I would have thought that as $A$ and $B$ are two operators, the indexing for the subsequences could be different and that we should be considering, say, $(Ax_{n(k)})$ and $(Bx_{n(j)})$ as the convergent subsequences.

So why is this not the case and why does the same indexing apply to both operators when specifying the convergent subsequences?

Answer 1

Such manipulations are often left implicit in more advanced textbooks. You can choose a subsequence $x_{n_k}$ for the operator $A$ such that $Ax_{n_k}$ converges. Then you can choose a subsequence of $x_{n_k}$ (a bounded sequence, being a subsequence of a bounded sequence) $x_{n_{k_l}}$ such that $Bx_{n_{k_l}}$ converges. Since $Ax_{n_{k_l}}$ is a subsequence of the convergent sequence $Ax_{n_k}$, it also converges so we have a sequence $x_{n_{k_l}}$ such that both $Ax_{n_{k_l}}$ and $Bx_{n_{k_l}}$ converge and this is your required sequence (renamed as $x_{n_k}$).