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My definition of $k$-space is:

Let be $(X,\tau)$ a topological subspace and $A\subset X$.

$A$ is a $k$-closed in $X$ if $\forall K\subset X$ compact it happens that $A\cap K$ is closed in $K$.

$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.

My question is the following:

I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.

Let be $(X,\tau)$ a $k$-space and $A\subset X$ closed, then $(A,\tau_{A})$ is a $k$-space.

Proof:

Let be $B\subset A$ a $k$-closed.

Then $\forall K\subset A$ compact it happens that $B\cap K$ is closed in $K$.

As $K\subset A \subset X$ and compactness is an absolute property, then $K$ is compact in $X$

$\Rightarrow$ $\forall K$ compact of $X$ it happens that $B\cap K$ is closed in $K$

$\Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.

Finally $\overline{B}^{A}=\overline{B}^{X}\cap A=B\cap A=B$

$\Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space

Obs: I mean absolute compactness for the following:

Theorem: Let be $(X,\tau)$ a topological space and $A\subset X$, then $B\subset A$ is compact in $A$ if and only if $B$ is compact in $X$.

AmottX
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  • This question was answered here https://math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake. – Oliver Jones Jul 20 '18 at 00:10
  • Your statement "$\forall K$ compact of $X$ it happens that $B\cap K$ is closed in $K$" does not seem to follow from anything stated previously. – bof Jul 20 '18 at 00:25
  • what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration.

    But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $B\cap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.

     – AmottX Jul 20 '18 at 03:27
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    Suppose $K$ is a compact subset of $X,$ how do you know that $B\cap K$ is closed in $K?$ If $K\subset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $K\not\subset A?$ – bof Jul 20 '18 at 03:32
  • Now, if $A$ happens to be *closed, then $K\cap B$ is compact so you can use that; but you say you are not* assuming that $A$ is closed. – bof Jul 20 '18 at 03:36
  • ok I already understood your point of taking an arbitrary compact $K$ of $X$ that does not necessarily $K\subset A$.

    Now only one last question. Assuming that $A$ is closed in $X$, what I can prove is that $A\cap K$ is compact in A and use the hypothesis that $B$ is a $k$-closed in $A$ and finally conclude that $B$ is a $k$-closed of $X$, is not it?

     – AmottX Jul 20 '18 at 04:31

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