Example of a subspace of a $k$-space which is not itself a $k$-space. I would like to know an example that shows that being a $k$-space is not a hereditary property.
My definition of $k$-space is:
Let be $(X,\tau)$ a topological space.


*

*For $A\subset X$, $A$ is $k$-closed in $X$ if for each $K\subset X$ compact it happens that $A\cap K$ is closed in $K$.

*$X$ is a $k$-space if every $k$-closed subset of $X$ is closed in $X$
I already proved that being a $k$-space is inherited by closed subspaces, but I would like to know an example of a subspace of a $k$-space which is not a $k$-space.
 A: Take $X = \{ ( \frac{1}{m} , \frac{1}{n} ) : n,m \geq 1 \} \cup \{ ( \frac{1}{m} , 0 ) : m \geq 1 \} \cup \{ (0,0) \}$, and give it the topology described as follows:


*

*for $m,n \geq 1$, the point $( \frac{1}{m} , \frac{1}{n} )$ is isolated;

*for $m \geq 1$ the basic open neighborhoods of $( \frac{1}{m} , 0 )$ are of the form $\{ ( \frac 1m , 0 ) \} \cup \{ ( \frac 1m , \frac 1n ) : n \geq k \}$ for $k \geq 1$;

*the basic open neighborhoods of $( 0,0 )$ are of the form $\{ ( 0 , 0 ) \} \cup \bigcup_{m \in A} U_m$ where $A \subset \mathbb{N}$ is cofinite, and $U_m$ is a basic open neighborhood of $( \frac 1m , 0 )$ for each $m \in A$.


Let $Y$ be the subspace $X \setminus \{ ( \frac 1m , 0 ) : m \geq 1 \}$.
The space $X$ is a sequential Hausdorff space, and so it is a k-space.  (If $A \subset X$ is not closed there is a sequence $( x_n )_n$ of distinct points of $A$ converging to some $x \in \overline{A} \setminus A$. Letting $K = \{ x_n : n \geq 1 \} \cup \{ x \}$ note that $K$ is compact, however $A \cap K = \{ x_n : n \geq 1 \}$ is not closed in $K$.)
To see that $Y$ is not a k-space, note the following.


*

*The set $Z = Y \setminus \{ (0,0) \}$ is not closed.

*The only compact subsets of $Y$ are the finite sets.
To begin, for each $m \geq 1$ let $A_m = \{ ( \frac 1m , \frac 1n ) : n \geq 1 \}$.
If $A \subseteq Y$ is infinite, then either there is some $m \geq 1$ such that $A \cap A_m$ is infinite, or there are inifnitely many $m \geq 1$ such that $A \cap A_m$ is nonempty.  In the first case note that $\{ Y \setminus X_m \} \cup \{ \{ \frac 1m , \frac 1n \} : n \geq 1 \}$ is a cover of $A$ with open subsets of $Y$ with no finite subcover.  In the second case, setting $B = \{ m \geq 1 : A \cap A_m \neq \emptyset \}$ and for each $m \in B$ picking some $n_m \geq 1$ such that $( \frac 1m , \frac 1{n_m} ) \in A \}$ it follows that $\{ Y \setminus \{ ( \frac 1m , \frac 1{n_m} ) \} : m \in B \} \} \cup \{ \{ ( \frac 1m , \frac 1{n_m} ) \} : m \in B \}$ is a cover of $A$ by open subsets of $Y$ with no finite subcover.
As the compact subsets of $Y$ are the finite subsets (and since $Y$ is Hausdorff) it is clear that $Z \cap K$ is closed in $K$ for each compact $K \subset Y$, meaning that $Z$ is a k-closed subset of $Y$.

It turns out that that the only Hausdorff spaces which are "hereditarily k-spaces" are the Fréchet spaces.  (A topological space $A$ is Fréchet if for each $A \subseteq X$ and $x \in \overline{A}$ there is a sequence in $A$ converging to $x$.)
