Prove that the set $A=\left \{ \frac{1}{n}\ |\ n\in \mathbb{N} \right \}$ is not compact.

Question

I am trying to prove this by showing that $A$ has no finite subcover. I have shown that $A$ is contained in an open cover, but am stuck to proceed forward.

My attempt:

Suppose we have a collection of open sets, $U = \left \{ (\frac{1}{n},2):n\in\mathbb{N} \right \}$. Choose arbitrary $\frac{1}{n}$ from $A$ and for all $n$, we see that $\frac{1}{n+1} <\frac{1}{n}<2$. Hence, $A$ is contained in $\bigcup_{n\in \mathbb{N}} (\frac{1}{n},2)$ so $U$ is an open cover of $A$.

Answer 1

Now we proceed using proof by contradiction:

Suppose that we have such a finite subcover, then we have $A\subseteq\bigcup_{n\in I} (\frac{1}{n},2)$ with $I=\{n_1,\ldots,n_k\}$ (finitely many indices), so we take $m=\max\limits_{1\leq i\leq k}n_i$ and note that $A\subseteq\bigcup_{n\in I} (\frac{1}{n},2)\subseteq \bigcup_{n=1}^m (\frac{1}{n},2)=(\frac{1}{m},2)$, but we see that $\frac{1}{m+1}\in A$ and $\frac{1}{m+1}\not\in (\frac{1}{m},2)$, which is a contradiction.

Answer 2

Continue with your work. Take any finite subcover $V$ of $U$: since $U$ is the union of an increasing sequence of open sets $(\frac1n,2)$, $V = (\frac1m,2)$ for some integer $m$. Choose $n = m+1$ so that $\frac1n \in A$ lies outside $V$. Therefore, no finite subcover of $U$ can cover $A$.