# Is a complement finite topological space compactly generated?

I'm using the following definition: A topological space $X$ is called compactly generated if it verifies: Any subset $U$ of $X$ is open iff for any Hausdorff compact space $Y$ and continuous map $f:Y\mapsto X$, $f^{-1}(U)$ is open. I find the requirement of Hausdorffness on the source spaces make it rather hard to check if a non-Hausdorff topological space is compactly generated directly through the definition. Thus I come about this question:

Let $C$ denote a countable infinite set endowed with complement finite topology (that is, the set of closed sets consists of $C$ itself and all its finite subsets). Then is $C$ compactly generated?

Let $C$ be as in your definition and let $\tilde{C}$ be $C$ (as set) equipped with the discrete topology. Denote with $K$ the one point compactifiction of $\tilde{C}$ this is a compact Hausdorff space. Let $c \in L\subset C$ be some element in a non-open set $L$ and define $f:K\to C$:

$$f(x)=\begin{cases} x \qquad \text{ if } x \in C\\ c \qquad \text{ if } x=\infty \end{cases}.$$

This is continous since if $U\subset C$ doesn't contain $c$ you have that $f^{-1}(U)=U$ which is open in $K$. If $U$ contains $c$ you have that $K\setminus f^{-1}(U)$ is compact in $C \subset K$ (it is just finitely many elements) but this means that $f^{-1}(U)$ is open in $K$ by definition of the open sets in the one-point compactification.

Now there are two cases: $L$ is infinite, than $C\setminus L$ is also infinite and therefore $K\setminus f^{-1}(L)$ is infinite and so it is not compact. Therefore $f^{-1}(L)$ is not open.

If $L$ is finite the same argument holds. So we have constructed a function such that a non-open set has a non-open preimage. Therefore $C$ is compactly generated.

• I think it's necessary to check that $f^{-1}(U)$ is open for any $f$ with compact Hausdorff domain, not a certain one, isn't it? – Censi LI Jun 15 '16 at 21:34
• Hmm, it seems you're right. I need to think about it for a second. – Maik Pickl Jun 15 '16 at 21:40
• @CensiLI I edited the answer. – Maik Pickl Jun 15 '16 at 21:46
• Very heuristic, than you very much. – Censi LI Jun 15 '16 at 21:54
• @StefanHamcke An infinite subset of $K$ is compact iff it contains $\infty$. And for any specific $L$, we can always arrange a $f$ such that $f^{-1}(L)$ doesn't containt $\infty$. – Censi LI Jun 16 '16 at 0:25

Since $C$ is a countable space, there are only countably many finite subsets. This means there are only countably many open sets, so $C$ is first-countable.
Any first-countable space is sequential, i.e. a subset $A$ is closed if $f^{-1}(A)$ is closed for any map $f$ whose domain is $\Bbb N^*$, the one-point-compactification of $\Bbb N$.
Now if $f^{-1}(A)$ is closed for any map $f$ from a compact Hausdorff domain to $C$, then $f^{-1}(A)$ is closed for any map $f: \Bbb N^* \to C$, hence $A$ is closed in $C$. And that means $C$ is compactly generated.

It would be interesting to ask whether an uncountable space with the cofinite topology is compactly generated.

Edit: An uncountable space $X$ with the cofinite topology is still sequential: For any non-closed set $A$ and any point $x\notin A$, there exists a sequence $(x_n)_n$ in $A$ converging to $x$ (Take any sequence without repetitions). That means $A$ is not sequentially closed, it does not have closed preimage under the map $\Bbb N^* \to X$ which sends $n$ to $x_n$ and $\infty$ to $x$.

• An uncountable set with the cofinite topology is still sequential: it's easy to see a set is closed iff it is sequentially closed. – Eric Wofsey Jun 16 '16 at 0:01
• @EricWofsey: I was just thinking about this when you posted your comment. Specifically about whether for a point $x$ in the boundary of a set $A$ there exists a sequence in $A$ converging to $x$ (which implies that the space is sequential). It seems to be true, mainly because any infinite sequence without repetitions converges to any point. – Stefan Hamcke Jun 16 '16 at 0:11
• A great overkilled proof. Thank you very much. – Censi LI Jun 16 '16 at 0:43