I cannot for the life of me seem to understand this proof. Here is my attempt.
Proof:
"==>": If $S$ is complete, every Cauchy sequence in $S$ converges to a point $x \in S$. Then $x$ is a limit point of $S$. Thus $S$ is closed.
"<==": If S is closed, then it contains all limit points. So $x \in S$. If $x$ is a limit point, then there is a sequence $a_n$ (with $a_n$ an element of S for all natural numbers) and $a_n$ converges to $x$. So the sequence is Cauchy and converges to a point in S. Thus, S is complete. QED.
What is wrong with my proof? Thanks for the help.
Edit: my updated proof for the direction "<==" is as follows:
Let $a_n$ be Cauchy in S and so $a_n$ converges to a point $x \in R$ (by Completeness of R). (We want to show $a_n$ converges to a point in S). So x is a limit point of S. Since, by assumption, S is closed and so it contains all limit points, this implies $x \in S$. Thus, every Cauchy sequence in S converges to a point in S. So S is complete. Is this correct now?
Best Answer
The main problem is that you’ve not really sorted out exactly what you need to prove. For the first part, you want to show that $S$ is closed, so you must let $x\in\Bbb R$ be an arbitrary limit point of $S$ and prove that $x\in S$.
For the second part you must show that every Cauchy sequence in $S$ converges to a point of $S$. Your argument clearly cannot be doing that, because you did not start with an arbitrary Cauchy sequence in $S$: you started with what you’re supposed to be finding, namely, a limit.
I’ve been a bit wordy in both arguments, but when you’re just starting to write proofs, it’s better to give more rather than fewer details, and I wanted to make clear exactly what the logic of the arguments is.