# Proof that every bounded sequence in the real numbers has a convergent subsequence

First, I realise the proof of this can be found in, say, Theorem 3.6 of Principles of Mathematical Analysis by W. Rudin: goes something like if the range of a sequence is finite, then the result is 'obvious', if not, you apeal to (Theorem 2.37) the fact that any infinite subset of a compact set has a limit point in said compact set (and then this in turn follows pretty much from the definition of compactness), and that by taking balls around the limit point in the range of the sequence that get smaller and smaller around this limit point you can construct some convergent subsequence, since every neighboorhood of a limit point has infinetly many points of the range (Theorem 2.20). That's fine.

I had though, to provide this proof recently, and at the time I didn't had the result of this Theorem 2.37 in mind. What I did was to reason that this result had to depend on the LUB property of the Reals (since the result is not true in, say, the rationals) and tried to construct some convergent subsequence from that. So, after all this rambling (sorry about that), my question really is if the general idea of what I did, which is the something like the following, works:

Let $\{p_n\}$ be some bounded sequence of real numbers. If the range of $\{p_n\}$ is finite, then the result is 'clear'. Suppose then it's infinite. Consider the $\sup(\mathrm{range}\{p_n\})$; if it is not in the sequence, then there is a convergent subsequence to it (because of how sup's are defined and something like Theorem 2.20), and we are done. Then suppose it's in the sequence. Then either there is some subsequence that converges to it and we are done, or not, in which case it's an isolated point (because of something like Theorem 2.20) and then define $y_1$ to be $\sup(\mathrm{range}\{p_n\})$. Having defined $y_1,\ldots ,y_{k-1}$, check if $\sup(\mathrm{range}\{p_n\}\setminus \{y_1,...,y_{k-1}\})$ is in the sequence; if not, we are again done, and if it is and there is no convergent subsequence to it, then define it to be $y_k$. Then either we found a convergent subsequence, or $y_1,y_2,\ldots ,y_n, \ldots$ is a monotonic decreasing convergent subsequence, since $\{p_n\}$ is a bounded sequence.

Is this just garbage, could it be improved, or is this fine?

Your proof is fine. Note that you are essentially constructing $\inf\{\,\sup\{\,p_n\mid n\ge k\,\}\|k\in\mathbb N\,\}$, i.e. $\limsup_{n\to\infty} p_n$.