0
$\begingroup$

Let $k\in \mathbb{N}$ define $E_k = \{ \frac{ 1}{ k } + \frac{ 1 }{ n} : n \in \mathbb{N} \}$ let $E = \cup _{k=1}^{\infty} E_k$

Is the set $E$ defined above a compact set it is bounded below by 0 and above by 2 so if it is closed we are done.

$\endgroup$
  • 3
    $\begingroup$ Remember that compact sets in a metric space will have no sequences that do not have a convergent subsequence. Can you construct a sequence in $E$ that fails to have a convergent subsequence? $\endgroup$ – Neil May 22 '17 at 14:34
  • $\begingroup$ @Neil If we add the zero to E then it will be compact? $\endgroup$ – Ameryr May 22 '17 at 14:44
  • 1
    $\begingroup$ I believe so, yes. $\endgroup$ – Neil May 22 '17 at 14:48
4
$\begingroup$

It is not closed. $0$ is a limit point, since $2/k \in E_k$ for $k=1,2,\ldots$. But the set $E$ does not contain $0$, as it consists of strictly positive numbers. Hence $E$ is not compact.

$\endgroup$
3
$\begingroup$

For $k=n$, $u_n=\frac{2}{n}\in E$, converges to $0$, but $0\notin E$.

$\endgroup$

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.