3

Let $X$ be a topological space. It is known that if $A\subset X$ is finite then $A$ is compact. I want to know what condition must be added such that the converse holds, that is if $A$ is compact then $A$ is finite. I think that this condition holds whenever $X$ is discrete space.

What are other conditions such that every compact set is finite?

Is it true that if every compact space $X$ is finite then $X$ must be discrete space?

lolo
  • 101
  • A compactly generated $T_1$ space $X$ with this property must be discrete: If $A$ is any subset, then $A\cap K$ is finite, thus closed, whenever $K$ is compact, so $A$ is closed then. – Stefan Hamcke Sep 08 '15 at 19:23
  • So for example every first-countable $T_1$ space with this property is discrete. A first-countable space with this property which is not discrete would be $\Bbb N$ with the topology having as basis the sets ${k\mid k<n}, n\in\Bbb N$. – Stefan Hamcke Sep 08 '15 at 19:25

1 Answers1

0

The co-countable topology on an uncountable set $X$ also has this property. (The nonempty open sets are those $U \subseteq X$ such that $X \setminus U$ is countable.)

  • Proof. If $A \subseteq X$ is infinite, take a countably infinite subset $\{ x_i : i \in \mathbb N \}$ of $A$. Then for each $j$ the set $U_j = X \setminus \{ x_i : i \neq j \}$ is an open set, and $\bigcup_j U_j \supseteq A$. Note that $x_i \in U_j$ iff $i = j$. From this it follows that given $j_1, , \ldots , j_n$ we have $U_{j_1} \cup \cdots \cup U_{j_n} \not\supseteq A$.
Lääne-Viru
  • 6,918
  • Well co finite topology should also do the job. –  Sep 08 '15 at 16:54
  • @Rememberme The co-finite topology on an infinite set is compact. (Every nonempty open set omits only finitely many points, so you'd need only finitely more to cover the space.) So It can't be an example of a space where only finite subsets are compact. – Lääne-Viru Sep 08 '15 at 18:09