Compact set on $\mathbb{R}$ with this topology generated from the basis set $B=\{(–a,a)\ s.t.\ a \in \mathbb{R}\}$

Question

Let A be a subset of a topological space $X$, and let $O$ be a collection of subsets of $X$.
(i) The collection $O$ is said to cover $A$ or to be a cover of $A$ if $A$ is contained in the union of the sets in $O$.
(ii) If $O$ covers $A$, and each set in $O$ is open, then we call $O$an open cover of $A$.
(iii) If $O$ covers $A$, and $O'$ is a subcollection of $O$ that also covers A, then $O'$ is called a subcover of $O$.(Topology: Pure and Applied, Adams 206)

In the example/question I am asking, the topological is being generated by $B=\{(–a,a)\ s.t.\ a \in \mathbb{R}\}$. Thus, $(-1,1)$ would have a finite subcover as would a closed interval, say $[-3,3]$.

But what about a half open interval like $(1,4]$ or $[–3,1)$? Would they be compact and have a finite subcover?

Answer 1

Supppose a set $A$ has a maximum $M$ and a minumum $m$, and $O_i , i \in I$ is a cover for $A$. Then $m \in O_i = (-a_i,a_i)$ for some $a_i \in \mathbb{R}$,as it is in $A$ so needs to be covered. SO $m < a_i$. Similary $M \in (-a_j,a_j)$ for some $a_j$. One of these is the bigger set, say $a_j < a_i$, then $ A \subseteq(-a_i, a_i)$, because $x \in A$ implies $x \le M < a_i$, and $x \ge m > -a_j > -a_i$.

So all such $A$ with max and min are compact in this topology, but this kind of argument also works for more sets.

If however $A$ is unbounded below and unbounded above, then $(-n,n), n \in \mathbb{N}$ is a cover for $A$ that has no finite subcover (note that a finite subcover can be reduced to one element). So a compact $A$ needs to be bounded, and so $M=\sup(A)$ and $m =\inf(A)$ exist both. If the larger one of these (in absolute value)is not in $A$, we can also find a cover without a finite subcover. Mull this over a bit...

Answer 2

Necessary and sufficient conditions for non-empty $A$ to be compact in this topology: $$(1).\;A \text { is bounded, and}$$ $$(2).\; |\inf A|\geq\sup A\implies \inf A=\min A ,\quad \text {and}$$ $$(3).\; |\inf A|\leq \sup A \implies \sup A=\max A.$$

For example: $A=[-1,2)$ then $\{(-2+2^{-n}, 2-2^{-n}): n\in \mathbb N\}$ is an open cover of $A$ with no finite sub-cover so $A$ is not compact.

Another example: $A=(-1,2]$ or $A=(0,2] . $ If $C$ is an open cover of $A,$ there exist $s\in C$ with $2\in s$,and there exists $(-b,b)\in B$ with $2\in (-b,b)\subset s\in C.$ This requires $b>2$ so the one-member sub-family $\{s\}$ covers $A$, because $s\supset (-b,b)\supset [-2,2]\supset A.$ So $A$ is compact.