0
$\begingroup$

In $\mathbb{R}$ with the usual metric, it holds that, for any $F \subset \mathbb{R}$ non-empty and bounded above, sup $F$ $\in \overline{F}$.

Show that there exists a sequence $S := (x_n)_{n \in \mathbb{N}}$ in $F$ with $x_n \rightarrow$ sup $F$ w.r.t. the usual metric.

I'd like to know if I'm approaching this problem correctly.

By definition, $\overline{F}$ = $F \bigcup F'$. Now, if sup $F \in \overline{F}$, then sup $F \in F$, sup $F \in F'$, or both.

If sup $F \in F'$, then by definition of $F'$ sup $F$ is an accumulation point. So, there must be some sequence that approaches sup $F$.

If sup $F \in F$, then this means sup $F$ is the maximum of $F$. So, $\forall f \in F$, $f \le$ sup $F$. Then we can take all terms of $F$ and construct that as a sequence. Since all the terms of this sequence are $\le$ sup $F$, the sequence approaches sup $F$.

Is this the correct way to approach this problem? I think I showed the sup $F \in F'$ part right, but I'm not very sure if the other case is true. I just tried to use the fact that sup $F \in F \implies$ sup $F$ is the maximum to try to "construct" a sequence with all the terms of $F$. Is this a valid approach? Or is there a simpler way to prove the problem. Thank you for your help.

$\endgroup$
1
$\begingroup$

I think it is more useful to use a more general approach, if you already know that $\sup E \in \overline{E}$.

It suffices to show that for every $x \in \overline{E}$, there exists a sequence $(x_n)_n$ such that $x_n \to x$.

This is easy:

By definition, $\forall \epsilon > 0: (x- \epsilon , x + \epsilon) \cap E \neq \emptyset$, so letting $\epsilon:= 1/n (n = 1,2, \dots)$, we can find a sequence $(x_n)_n$ in $E$ with $x_n \in (x-1/n, x+1/n)$ for all $n \geq 1$ meaning that $|x_n - x| < 1/n$ for all $n\geq 1$, and hence $x_n \to x$

$\endgroup$
  • $\begingroup$ Guess I should get used to thinking more generally than breaking the problem down into specifics. Thank you! $\endgroup$ – Max Mar 4 '18 at 10:45
  • $\begingroup$ You are welcome! Note that the axiom of choice was used when picking elements out of every set in the collection $\{(x-1/n,x+1/n) \mid n \geq 1\}$ $\endgroup$ – user370967 Mar 4 '18 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.