Are all elements of a bounded sequence also elements of some convergent subsequence (proof suggestion)?

Question

I'm trying to prove the following proposition (no idea if it's actually true or not):

All elements of a bounded sequence are also elements of some convergent subsequence.

The idea I thought about, laid out informally, is:

Let $a_n$ be a bounded sequence, so by the Bolzano-Weierstrass theorem it has a convergent subsequence $a_{n_k}$.

Since $a_{n_k}$ is also bounded, by using Bolzano-Weierstrass again, we can get a subsequence $a_{n_{k_i}}$, which is also convergent (and to the same limit as $a_{n_k}$, of course) . In that manner we can generate an infinite amount of convergent subsequences of $a_n$, and the set of these subsequences is countable.

We can take any element of the set $\left\{a_n\:|\:n\in N,\:a_n\notin a_{n_k}\right\}$, which is also countable, and "insert" each individual one of them before the first element of a different convergent subsequence (there are infinitely many), and so it turns out that all elements of $a_n$ are also elements of some convergent subsequences, as requested.

As I said, I don't even know if this proposition is correct, it just bothered me so I tried to ascertain its validity. I don't know if this proof is reliable, becausen infinity is a tricky beast... reviews and corrections would be very helpful :).

Answer 1

Your proof is nearly correct (two minor mistakes) but overcomplicated.

• "and the set of these subsequences is countable": no, it has the cardinality of the continuum.
• '"insert" each individual one of them before the first element of a different convergent subsequence' is the good idea but
  • not before the first element, because its index in the sequence $(a_n)$ might be smaller than the index of the element you want to insert;
  • you do not need to do this with distinct subsequences of your convergent subsequence $(a_{n_k}).$ You can do this directly with that subsequence. Alternatively:

For any $n\in\Bbb N,$ let $k_0\in\Bbb N$ be such that $n_{k_0}>n.$ Then, the subsequence whose first term is $a_n$ and the following terms are $(a_{n_k})_{k\ge k_0}$ is convergent.