Z compact with another topology?

Question

Consider $\mathbb{Z}$ equipped with a topology generated by the basis of the form $(-k,k)$, is a finite set, say (-3,3), compact? Is $\mathbb{Z}$ compact?

I think a finite set is compact since you can just find a finite (in fact singleton) subcover, that is the interval that includes all the points in the finite set. For the second part, $\mathbb{Z}$ is not for the same reason that it is not compact in the standard topology.

Can someone please check if what I am thinking is correct? Thanks

A finite set is compact in $\mathbf{any}$ topology. – Nitrogen Mar 23 '16 at 04:20 — Nitrogen, Mar 23 '16 at 04:20

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

0

Let $(X, \tau)$ be a topological space. Suppose that $A \subseteq X$ so that $|A|<\infty$. Suppose that $\mathcal{U}$ is an open cover of $A$. Then we go ahead and choose $U_a$ so that $a \in U_a$. There are at most finitely many $U_a$. Then $A^{\prime}=\{U_a \mid a \in U_a\}$. But then there is a finite sub-collection so that $A \subseteq A^{\prime}$, hence $A$ compact. This is more general. Any finite set is compact.

I'm not sure what you meant by "singleton sub-cover," but the set of $\{x\} \subseteq X$ need not be open. See here for further discussion: Are Singleton sets in $\mathbb{R}$ both closed and open?

edited Apr 13 '17 at 12:20

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answered Mar 23 '16 at 04:39

Andres Mejia

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AC is unnecessary. You have a finite amount of choices. – Aloizio Macedo Mar 23 '16 at 04:44
Is this true? What if the chosen subcover is infinite? – Andres Mejia Mar 23 '16 at 04:45
Point taken, you are correct. Will edit – Andres Mejia Mar 23 '16 at 04:46
So is it because we cannot use this reasoning that Z is not compact in this space? – user325221 Mar 23 '16 at 14:54
What would it take to "cover" $\mathbb{Z}$? For every cover, is there a finite subcollection that suffices? – Andres Mejia Mar 23 '16 at 15:53

Z compact with another topology?

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