# Is this set a compact set

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.

• 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? – Neil May 22 '17 at 14:34
• @Neil If we add the zero to E then it will be compact? – Ameryr May 22 '17 at 14:44
• I believe so, yes. – Neil May 22 '17 at 14:48

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