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Given a space $X$ and a collection of subspaces $X_\alpha$ whose union is $X$, these subspaces generate a possibly finer topology on $X$ by defining a set $A\subset X$ to be open iff $A\cap X_\alpha$ is open in $X_\alpha$ for all $\alpha$. In case $\{X_\alpha\}$ is the collection of compact subsets of $X$, we write $X_c$ for this new compactly generated topology.

Given topological spaces $X$ and $Y$, do $X\times Y$ and $X_c\times Y$ have the same compact subsets?

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    $\begingroup$ What is $S_c$ for a set, $S$? $\endgroup$ – Adam Hughes Jan 29 '15 at 20:23
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    $\begingroup$ @AdamHughes Probably $X_c$ denotes the topogical space that has the same underlying subset as space $X$ and is equipped with the topology inherent with compact subsets: $F$ is closed in $X_c$ iff $F\cap C$ is closed for each compact $C\subseteq X$ $\endgroup$ – drhab Jan 29 '15 at 21:03
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Here are some hints: (Whenever I speak of a map betweens spaces, I'll mean a continuous function.) Note that the compact subsets of $X$ are exactly the images $t(C)$ of maps $t:C→X$ where $C$ is a compact space (call such a map a test map). Then $X_c$ has the final topology for all test maps to $X$, which are the same as the test maps to $X_c$, so $X_c$ is compactly generated, let's call this a c-space. For a $c$-space $X$, a function $f:X→Y$ is a map if $ft$ is a map for each test map $t:C\to X$. Now it's easy to show that for a $c$-space $X$, a map $X→Y$ is also a map $X→Y_c$.

Since $X$ and $X_c$ have the same test maps, they also have the same compact subsets. With the remarks above, can you establish a sequence of maps $$(X\times Y)_c\to (X_c\times Y_c)_c\to X_c\times Y_c\to X_c\times Y\to X\times Y$$ and conclude that all these spaces have the same compact subsets?

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