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I am using the definition of compactly generated space from The Category of CGWH Spaces, which is

In $\mathbf{Top}$, a $k$-closed subset $Y\subset X$ is a set such that $u^{-1}(Y)$ is closed in $C$ for any $u: C\to X$ where $C$ is compact Hausdorff.

A space is compactly generated if all $k$-closed subsets are closed.

locally compact means every point has a local base of compact sets.

This is different from the definition from Wikipedia

So, is any compact space compactly generated?

And, is any locally compact space compactly generated?

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For your first question, if $X$ is compact Hausdorff, consider taking $C=X$ and $u$ the identity. –  Nate Eldredge May 6 at 5:12
Yes, I know this. But I want to know what about non-Hausdorff space –  Minghao Liu May 6 at 5:59
The term compactly generated is used differently by different authors. Sometimes is denotes a space where a subset is closed if it intersects each compact subset in a closed set, let's call these c-spaces. Other times it denotes a space as you describe it, but these are also called k-spaces. Both compact and locally compact spaces are c-spaces. So you are actually searching for a space that is a c-space but not a k-space. –  Stefan Hamcke May 19 at 14:48
It seems that both your question and this require a space with many compact subsets which are not images of compact Hausdorff spaces. –  Stefan Hamcke May 19 at 14:49
By the way, is there any explicit example of a compact space, even non-T₀ one, to where a Hausdorff compact can’t be surjectively mapped? –  Incnis Mrsi yesterday

1 Answer 1

It seems the following.

We shall use this question by Amathstudent and its answer by Brian M. Scott.

We prove that each weak Hausdorff compactly generated $T_1$ space is $KC$. Since there is a weak Hausdorff compact $T_1$ space, which is not $KC$ (see the space $\Bbb Q^*\times\Bbb Q^*$ in the answer by Brian M. Scott), this space is not compactly generated.

So, let $X$ be a weak Hausdorff compactly generated $T_1$ space and $Y$ be a compact subset of $X$. We claim that $Y$ is $k$-closed subset of $X$. Indeed, let $C$ be a compact Hausdorff space and $u: C\to X$ be a continuous map. Since the space $X$ is weak Hausdorff, the set $u(C)$ is closed in $X$. A set $u(C)\cap Y$ is compact as a closed subspace of a compact space $Y$. By Lemma 1, the space $u(C)$ is Hausdorff. So the set $u(C)\cap Y$ is closed in the space $u(C)$. Since the set $u(C)$ is closed in $X$, the set $u(C)\cap Y$ is closed in the space $X$ too. Since the map $u$ is continuous, the set $u^{-1}(Y)= u^{-1}(Y\cap u(C))$ is closed in $C$. Since the space $X$ is compactly generated, the set $Y$ is closed in $X$. Hence $X$ is a $KC$-space.

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Can you show a k-closed subset in your example that isn’t closed? Is it the diagonal? –  Incnis Mrsi yesterday
@IncnisMrsi The diagonal is not closed as remarked Brian M. Scott, but it should be $k$-closed as I claimed in the old version of my answer. –  Alex Ravsky yesterday
@AlexRavsky Brilliant. Just a minor note: weakly Hausdorff spaces are already $T_1$ (any point is closed being the image of some constant function $u\colon K\to X$). What about the second question about locally compact spaces? In $\mathbb{Q}^*\times\mathbb{Q}^*$ no point has a local base of compact sets so it does not disprove the conjecture "locally compact implies compactly generated". –  johndoe yesterday
Incidentally, your result shows that not every compact space is homeomorphic to a quotient of some locally compact Hausdorff space (compare this to the fact that every space is homeomorphic to a quotient of some Hausdorff space). –  johndoe yesterday
@IncnisMrsi: The diagonal in $\Bbb Q^*\times\Bbb Q^*$ is indeed $k$-closed. This can also be deduced as follows: Since $\Bbb Q^*\times\Bbb Q^*$ is weak Hausdorff, so is its $k$-ification $k(\Bbb Q^*\times\Bbb Q^*)$ (this equippes the space with the final topology with respect to all maps from compact Hausdorff spaces). But a $k$-space is weak Hausdorff iff the diagonal is closed, see Neil Stickland's paper CGWH. That means the diagonal is $k$-closed, but it's not closed as $\Bbb Q^*\times\Bbb Q^*$ is not Hausdorff. –  Stefan Hamcke yesterday

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