Compactness of a set : $A=\left\lbrace 2019,1010,\ldots,(n+2018)/n,\ldots \right\rbrace$ and $B=A \cup \left\lbrace 1 \right\rbrace$

Question

How can I show directly (that is, not using the compactness criterion in $\mathbb{R}$) that every open cover $G$ of $B$ admits a finite subcover?

I know that $A$ is not compact and I am aware that I would need a single example to prove it, any hint on how to find an open cover of $A$ that does not contain a finite subcover?

Answer 1

If I take some open cover of $B$, there is some (at least one) open set $U_1$ which covers $1$. It must also cover some neighborhood of $1$ (since it is an open set). Since it covers some neighborhood of $1$, it will also cover almost any member of $A$ (you need to explain why in your proof - this is the main issue). Thanks to that fact, you only need finitely more open sets from the cover in order to cover $B$.

Answer 2

We have $A=\{a_n\}_{n=1}^\infty$, note that $1<a_{n+1} < a_n$ and $a_n \downarrow 1$.

Let $b_n = {1 \over 2} (a_n+a_{n+1})$, the point half way between $a_n, a_{n+1}$.

Now take the open cover consisting of $(b_1,2020)$ and $(b_{n+1},b_n) $ for $n=1,...$.

Note that $A$ is not contained in any finite subcollection, hence $A$ is not compact.