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Does anyone know of an example of a topological space which is not compactly generated? I am using the definition in May's book "A Concise Course in Algebraic Topology." The definition is that a space $X$ is compactly generated if for any continuous map $g:K\to X$ from a compact Hausdorff space $K$, we have $g(K)$ is closed in $X$ (i.e. $X$ is "weak Hausdorff") and for any subset $A$ of $X$, $A$ is closed if an only if for any map $g:K\to X$, the preimage $g^{-1}(A)$ is closed in $K$ (i.e. $X$ is a "$k$-space").

Of course, any compactly generated space is at least $T_1$, as compact Hausdorff spaces are $T_1$, so I am really interested in an example which has $T_1$-separation (or even better, one which is Hausdorff) but which fails to be a $k$-space.

CWcx
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3 Answers3

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An example I learnt from Engelking's book:

let $X = \mathbb{R} \setminus \{1,\frac{1}{2},\frac{1}{3},\ldots\}$ in the subspace topology and let $Y = \mathbb{R}$ where we identify the positive integers $\mathbb{N}$ to a point, in the quotient topology. Then $X \times Y$ is not a $k$-space (or compactly generated).

See this blogpost or Engelking's chapter on $k$-spaces for the argument. As this space is even Tychonoff, compactly generated and $k$-space etc. all coincide.

Henno Brandsma
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I have read about $k$-spaces is Topology and Groupoids by Ronald Brown, where they are defined as spaces where a subset is closed if its preimage $f^{-1}(A)$ is closed for each continuous map $f$ from a compact Hausdorff space $K$ into $X$. I'm used to the notion of a compactly generated space as a space where a subset is closed if its intersection with each compact subset is closed. Some authors require the space to be Hausdorff, while May only requires it to be weak Hausdorff. Note that a $k$-space is always compactly generated, while for Hausdorff spaces both properties agree.

Whatever definition is used, it is not that difficult to find a space which is not compactly generated (and hence not a $k$-space), if you keep in mind that in a $T_1$ space the intersection with each compact set is automatically closed if every compact set is finite (A space where every compact set is finite is called anti-compact). So we just have to find an anti-compact $T_1$ space which is not discrete. There exists such a space, namely the space $X=(\mathbb R,\tau_{cc})$, the real line with the cocountable topology . It's easy to check that no infinite subset of $X$ is compact. But since this space is not discrete, it is not compactly generated.

I don't have an example of a Hausdorff space, though. But when you try to find such a space, keep in mind that the following spaces are always $k$-spaces:

  • spaces where each point has a compact Hausdorff neighborhood
  • sequential spaces
  • quotients of a topological sum of compact Hausdorff spaces

Edit: An example of a Hausdorff space which is not compactly generated is now presented in my answer to Spaces in which "$A \cap K$ is closed for all compact $K$" implies "$A$ is closed."

Community
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Stefan Hamcke
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Here is a very simple example. Let $X=\mathbb{N}\times\mathbb{N}\cup\{*\}$, where a subset $U\subseteq X$ is open if either

  • $*\not\in U$, or,
  • for all but finitely many $n\in \mathbb{N}$, $U$ contains all but finitely many points of $\{n\}\times\mathbb{N}$.

Clearly $X$ is Hausdorff (in fact, it is normal). It is easy to see that every infinite subset of $X$ contains an infinite closed discrete subset (choose either a subset contained in $\{n\}\times\mathbb{N}$ for some $n$ or a subset which contains only finitely many points in each $\{n\}\times\mathbb{N}$), so every compact subset of $X$ is finite. Thus the compactly generated topology on $X$ is just the discrete topology, so $X$ is not compactly generated.

(More generally, you could replace the second bullet above with "$U\in F$", where $F$ is any filter on $X$ which is not contained in the cofinite filter on any infinite subset of $X$. The case when $F$ is a nonprincipal ultrafilter is a useful example for many questions in pointset topology.)

Eric Wofsey
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