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Let $X$ denote a topological space. For any $A \subseteq X$, consider two possible conditions on $A$.

  1. $A$ is closed
  2. $A \cap K$ is closed, for all compact $K \subseteq X$.

If $X$ is Hausdorff, then clearly (1) implies (2). If $X$ is metrizable, then the converse holds; see here.

Questions.

  • What are some Hausdorff topological spaces in which (2) does not imply (1)?

  • For Hausdorff topological spaces: what weaker conditions than metrizability are known to imply that (2) implies (1)? Are there any interesting characterizations of those Hausdorff spaces in which (2) implies (1)?

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goblin GONE
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  • If (3) is the property $A\cap K$ is closed in $K$ for all compact $K\subseteq X$, then a space in which (3) implies (1) is called compactly generated. In a Hausdorff space (2) and (3) are equivalent, so you are asking for a Hausdorff space which is not compactly generated. – Stefan Hamcke Apr 19 '15 at 17:49
  • This answer gives an example of a Tikhonov space that is not compactly generated. All locally compact Hausdorff spaces and all sequential spaces are compactly generated. – Brian M. Scott Apr 19 '15 at 19:24
  • @StefanHamcke, interesting. – goblin GONE Apr 20 '15 at 06:00
  • Regarding your second question: All spaces where every point has a compact neighborhood, more general, spaces which are a quotient of a disjoint union of compact spaces are compactly generated. Also every first-countable space, more generally, every sequential space is compactly generated. – Stefan Hamcke Apr 21 '15 at 20:06

1 Answers1

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Let $Q$ be the space of non-negative rational numbers, $N$ the natural numbers including $0$, and $q:Q\to Y=Q/N$ the quotient map identifying $N$ to a single point. Note that $Y$ is a Hausdorff normal space since $Q$ is such a space and $q$ is a closed map.
Now look at $Y\times Q$. This space is Hausdorff. We will show that it's not compactly generated.
Let us call the image of $(n,n+1)$ under $q$ the cell $e_n$. Note that a compact subset $C$ of $Y$ can only intersect finitely many cells. Otherwise, we could construct a subset $S\subseteq C$ by picking a point $s_n$ from every non-empty intersection $e_n\cap C$. Every subset $T$ of $S$ then had a closed preimage under $q$, thus would be closed, hence $S$ would be an infinite subset of $C$ without a limit point. This also means that a compact set in $Y\times Q$ is contained in a finite number of "cylinders" $\overline{e_n}\times Q$.
We now construct a non-closed set $A$ in $Y\times Q$ such that $A\cap K$ is closed for every compact $K$.
Let $h=q\times\mathbf 1_Q$ and let
$$ A' = \left\{(a,b)\in Q^2\ \middle|\ \frac\pi{\lfloor a+1\rfloor} < b < \frac\pi{\lfloor a+1\rfloor} +\min(a-\lfloor a\rfloor,\lceil a\rceil-a)\right\}. $$ This set is closed and the preimage of $A=h(A')$. However, $A$ is not closed as $(0,0)$ is its limit point. If $K$ is compact, then $A\cap K=h(A'\cap h^{-1}(K))$. The set $A'\cap h^{-1}(K)$ is closed and contained in some $[0,m]\times Q$, and that means $A\cap K$ is the image of a closed set under the restriction $h|_{[0,m]\times Q}$. This map is closed, being the product of the perfect map $q|_{[0,m]}$ with the identity on $Q$ (where perfect is the terminology for a map which is closed and has compact fibers). So we see that $A \cap K$ is closed, even though $A$ is not closed.

Stefan Hamcke
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