# 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 $x_{n_k}$).
• The quoted text is not wrong nor misleading - indeed each bounded sequence $x_n$ has a subsequence $x_{n_k}$ for which both $Ax_{n_k}$ and $Bx_{n_k}$ converge. It just doesn't specify how you would find such a sequence so in this sense, this is a "very partial proof" whose details are left to the reader... Not that I'm trying to justify such "proofs" - from my point of view, your either don't bother to write a proof for this at all or, if you already do, you should add the extra line to make it less confusing. Aug 5, 2017 at 21:59