Same convergent subsequence for two compact operators? 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?

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 $x_{n_k}$). 
