Possible flaw in "proof" that a sum of two compact operators is compact

Find a convergent subsequence for $A$, then take a sub-subsequence for $B$. You can't necessarily intersect two subsequences, but you can easily take a sub-subsequence.

If you have two sequences, say $\{a_n\}$ and $\{b_n\}$, and sufficiently neat conditions, then both have a convergent subsequence indexed by the same same set of indices. Formally, I mean that there exists a sequenece of integers $\{n_1<n_2<n_3<\cdots\}$ such that both $\{a_{n_i}\}_{i=1}^\infty$ and $\{b_{n_i}\}_{i=1}^\infty$ both converge.

To guarantee this, you're going to need either

1. each of $\{a_n\}$ and $\{b_n\}$ lie in some compact set (even if they lie if different compact sets, it works); or
2. each of $\{a_n\}$ and $\{b_n\}$ are bounded so you can apply some kind of Bolzano-Weierstrass.

These amount to the same proof though.

In fact, you can get this "common subsequence trick" far more generally: instead of two sequences $\{a_n\}$ and $\{b_n\}$, you can have countably many sequences $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\ldots$. It's the same proof.

For two sequences it's actually very easy to write down. Let $\{a_n\}$ and $\{b_n\}$ be bounded, and say $\{a_n\}$ has convergent subsequence $\{a_{n_i}\}$. But then $\{b_{n_i}\}$ has a convergent subsequence because it is bounded --- say $\{b_{n_{i_k}}\}$ converges. But now $\{a_{n_{i_k}}\}$ is a subsequence of $\{a_{n_i}\}$, hence also convergent. Thus $\{n_{i_1} < n_{i_2} < n_{i_3} < \cdots\}$ is the required set of indices.

For the case with countably many sequences, you can imagine how to proceed. The hardest part is streamlining the notation.

Do you see how to apply this to prove that the sum of two compact operators is compact?

Consider the following:

We have the sequences $\{Ax_n\},\{B x_n\},$ and $\{(A+B)x_n\}$.

We begin by finding the a convergent subsequence of the first sequence; that is, we find a sequence of indices $\{n_i\}$ such that $\{A_{n_i}\}$ converges. Note that any subsequence of this convergent sequence must also converge.

We now find a convergent subsequence of the sequence $\{B x_{n_i}\}$ over the indices $n_{i_j}$ (which I will simply write as $n_j$, for convenience). Note that both $\{A_{n_j}\}$ and $\{B_{n_j}\}$ converge.

In fact you do it gradually:

Take $x_{n_j}$ such that $Ax_{n_j}$ converges, and then take a sub-subsequence $x_{n_{j_k}}$ such that $Bx_{n_{j_k}}$ converges

There are several ways to get rid of this difficulty:

• you can still extract o sequence (for $B$) from an extracted convergent subsequence for $A$.

• or, take an equivalent definition of a compact operator: this is an operator for which the image of a ball is compact.