TITLE: Is every compact space compactly generated?
QUESTION [8 upvotes]: 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?

REPLY [7 votes]: 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.