I am having some troubles understanding one bit about the converse of Heine-Borel which is the following statement:
Let $S$ be a subset of $\mathbb R^n$. Then the following three statements are
equivalent:a) $S$ is compact.
b) $S$ is closed and bounded.
c) Every infinite subset of $S$ has an accumulation point in $S$.
Here is the proof presented from the book that I am reading from I have problems in particular the implication $c) \Rightarrow b)$. I don't understand why that it is sufficient to prove that $x$ is the only accumulation point? But then according to that if I am understanding it correctly then every closed bounded set should have only 1 accumulation point which isn't true maybe I am confused here is picture of the proof. If someone could clarify the issues at which I am confused in that would be perfect.
Best Answer
Note that $x$ isn't the only accumulation point of the (very general) set $S$, it is the unique accumulation point of the Cauchy sequence $T$. And a Cauchy sequence can only have one accumulation point.
They want to show that $S$ is closed, and they do that by showing that $S$ contains any of its accumulation points. This means taking an arbitrary accumulation point $x$ of $S$ and show that it is contained in $S$. They do this by constructing $T$, an infinite subset of $S$ with $x$ as its only accumulation point (as a subset of $\Bbb R^n$).
Since $T$ is an infinite subset of $S$, there must be some accumulation point of $T$ contained in $S$, and there is only one possibility. Therefore $x\in S$, and since $x$ was an arbitrary accumulation point, we're done.