I'm trying to prove that if $X$ is compactly generated and $Y$ is T2 (Hausdorff) and locally compact then $X\times Y$ is compactly generated.

First it is clear that since both $X$ and $Y$ are T2 then $X\times Y$ is also T2 and so the first condition for "compactly generated" is fulfilled.

For the next condition, we need to show that: $$\forall B\subseteq X\times Y,\,\left[B\in Open(X\times Y)\Longleftrightarrow B\cap K\in Open(K)\forall K\in Compact(X\times Y) \right]$$

So let $B\subseteq X\times Y$ be given.

The direction $\Longrightarrow$ follows immediately due to the definition of the subspace topology, so there is nothing to prove.

For the $\Longleftarrow$ direction, assume $B\cap K\in Open(K)\forall K\in Compact(X\times Y)$, and the goal is to show $B\in Open(X\times Y)$.

My idea was to take an arbitrary point $(x,y)\in B$ and try to find $U\times V \in Open(X)\times Open(Y)$ such that $(x,y)\in U\times V \subseteq B$.

Because $Y$ is locally compact, $\exists K_y \in Compact(Y)$ such that $\exists V_y \in Open(Y)$ such that $y\in V_y \subseteq K_y$.

However, now I get stuck, because I don't know which compact set $K_x$ of $X$ to find so that $x\in K_x$ and $(x,y)\in K_x \times K_y$.

Because there is no obvious choice for a compact set, I'm not sure how to employ the input data.


A space $X$ is a $k$-space if it has the final topology with respect to all maps from compact Hausdorff spaces to it, in other words, if $A\subseteq X$ is closed if $t^{-1}(A)$ is closed in $K$ for every map $t:K\to X$ where $K$ is compact Hausdorff. Usually, a compactly generated space has the property that a subset is closed if it intersects every compact subspace in a closed set, so every $k$-space is compactly generated. For Hausdorff spaces, though, both definitions are the same.

It is useful to know that a space is a $k$-space (without $T_2$) precisely if it's a quotient of a disjoint union of compact Hausdorff spaces. This also implies that a space in which every point has a compact Hausdorff neighborhood, and this includes locally compact Hausdorff spaces, is a $k$-space.

So in your problem you have a Hausdorff $k$-space $X$ with a quotient map $p:\bigsqcup_i K_i\to X$, and you have a locally compact Hausdorff space $Y$ with a quotient map $q:\bigsqcup_j L_j\to Y$, where all $K_i$ and $L_j$ are compact Hausdorff. This gives you a surjective map $p×q:\bigsqcup_{i,j}K_i×L_j\to X×Y$, and the goal is to show that $p×q$ is a quotient map. But we can factor this map as $(p×1_Y)(1_{\bigsqcup K_i}×q)$, and each of these maps is quotient map, being the product of a quotient map with the identity map on a locally compact space.

  • $\begingroup$ Thanks for your answer. Is what you are saying equivalent to saying that X is compactly generated if it is homeomorphic to the direct limit of the collection of its compact subspaces? (a la Spanier alg. top. page 5) $\endgroup$ – PPR May 19 '15 at 9:30
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    $\begingroup$ @PPR: It is true that a space is compactly generated if and only if the canonical map from the colimit of the diagram of its compact subspaces to that space is a homeomorphism. $\endgroup$ – Stefan Hamcke May 19 '15 at 15:20

Some quotations from “General Topology” by Ryszard Engelking:

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