A Hausdorff $ k$-space is minimal $ KC$ if and only if it is compact.

Question

Atopological space is called $k$ - space if it has the property that any subset $S$ such that $ S \cap K $ is closed for all closed compact $ K $ is itself closed.

The bellow theorem comes from " THE FDS-PROPERTY AND SPACES IN WHICH COMPACT SETS ARE CLOSED." I have some problem with this theorem.

*Theorem:*A Hausdorff $k$-space is minimal $ KC$ if and only if it is compact.

Proof: The sufficiency is clear. For the necessity, let $(X, τ)$ be a non-compact space which satisfies the hypothesis of the theorem. Define a new topology σ on X as follows: $σ = \{U ∈ τ : a \not\in U \} ∪ \{U ∈ τ : a ∈ U , \quad \text{X - U is compact} \}$. Clearly (X, σ) is a compact and σ ⊂ τ. we claim that $(X, σ)$ is a $KC$-space.To this end, suppose that $ S ⊆ X$ is a compact subset of $(X, σ)$. It is clear that $cl_{σ}(S) ⊆ cl_{τ} (S) ∪ \{a\}$ . There are then two possibilities: If $S$ is a compact subset of $(X, σ)$ and

(a) if $a \not\in S$, then then by the preceding remarks, S is compact, and hence closed, in $(X, τ)$ and so $X - S $ is an open $σ$-neighbourhood of a. Thus $a \not\in cl_{σ}(S)$ and so $cl_{σ}(S) = cl_{τ} (S) = S$. then $cl_{σ}(S) = cl_{τ} (S) $.

(b): If on the other hand, $a ∈ S$, then $cl_{σ}(S) = cl_{τ} (S)$

and so if $S$ is not closed in $(X, σ)$, then it is not closed in $ (X, τ)$ either. Since $(X, τ)$ is a $k$-space, there is some compact set $C$ in $ (X, τ) $ such that $ C ∩ S $ is not closed in $ C$. Furthermore, if the chosen compact set $ C $ has the property that $a ∈ C$, then since $(X, τ)$ is Hausdorff, given $x ∈ cl_{τ} (C ∩ S) - (C ∩ S)$, we can find disjoint open neighbourhoods $ U, V $ of $x$ and $a$ respectively. Then

$ C - V$ is a compact subset of $(X, τ)$ with the property that $S ∩ (C - V ) $ is not closed in $ C - V $. Hence we have shown that it is possible to choose$ C$ so that $a \not\in C $.

Then $cl_{τ} (C ∩ S) ⊆ C $ is a closed, hence compact subset of $ (X, τ) $ which does not contain $ a$ and hence is also closed in $ (X, σ)$ . Thus $T = S ∩ cl_{τ} (C ∩ S)$ is a $σ$-closed subset of $S$ and hence is compact in $(X, σ)$. However, since $a \not\in T$ , it follows that $T_{τ} = T_{σ}$ and hence $T$ is compact in $(X, τ)$, a contradiction, since $x ∈ cl_{τ} (T ) - T$ .

But:

can you explain that : why he said that "

(1):$ C - V$ is a compact subset of $(X, τ)$ with the property that $S ∩ (C - V ) $ is not closed in $ C - V $ . Hence we have shown that it is possible to choose $ C$ so that $a \not\in C $.

(2): why $cl_{τ} (C ∩ S) ⊆ C $ is a closed?

(3) :$(X, τ)$ is $ T_{1}$, is it right to say that $(X, σ)$ is $ T_{1}$?

Thanks.

Answer 1

(A point you forgot to mention: $a \in X$ is just any (now fixed) point of $X$)

Point (3) is easiest to handle: Suppose $(X,\tau)$ is $T_1$; this is equivalent to the fact that all sets of the form $X \setminus \{x\}$ are in $\tau$. If $x \neq a$, then $X \setminus \{x\}$ contains $a$, and so is in $\sigma$ because its complement is $\{x\}$ which is compact (second part of the union defining $\sigma$). If $x = a$ then $X \setminus \{a\}$ is in $\tau$ and does not contain $a$ and so is in $\sigma$ because of the first part of that union. In all cases, all sets of the form $X \setminus \{x\}$ are also in $\sigma$, and so $(X, \sigma)$ is $T_1$ whenever $(X, \tau)$ is.

As to (1): In the proof, we have a compact (under $\sigma$) compact set $S$, that we assume is not closed in $(X,\sigma)$. We then know we can assume $a \in S$ (as the case $a \notin S$ was already handled) and in that case we know that $S$ has the same closure under both $\sigma$ and $\tau$ and so $S$ is also not closed in $(x, \tau)$.

We then apply that $(X, \tau)$ is a k-space to get a compact $C$ (in $(X,\tau)$, so $C$ is also closed there as this space is Hausdorff) with $S \cap C$ not closed in $C$, under the relative topology induced by $(X, \tau)$, of course.

The authors want this $C$ to have the property that $a \notin C$ as well (for the rest of the proof to work) so they show that if $a \in C$ (contrary to what they'd like) we can choose a new, smaller $C$ that does have the property, and this new $C$ is of the form $C \setminus V$, for the $C$ we start with and a neighbourhood $V$ of $a$, as shown in the proof.

Then in the remainder they assume that the $C$ (compact in $\tau$ and such that $S \cap C$ is not closed in $C$ under its $\tau$-topology) indeed has the property that $a \notin C$.

This is the reason of the remark "Hence we have shown that it is possible to choose $C$ so that $a \notin C$".

Then (2) follows: the sentence in the paper is: "Then $\operatorname{cl}_{\tau}(S \cap C) \subset C$ is a closed, hence compact subset of $(X, \tau)$, which does not contain $a$ and hence is also closed in $(X, \sigma)$".

Well, $\operatorname{cl}_{\tau}(S \cap C)$ is by definition closed in $(X, \tau)$, as it is the closure of a set, in this case $S \cap C$! It is a subset of $C$ because $C$ is already closed in $(X, \tau)$ (recall it is a compact set in a Hausdorff space), and the closure of $S \cap C$ under $\tau$ is the smallest $\tau$-closed set containing it (and $C$ is one of those closed sets). And it is in $\sigma$, because of the first part of the definition of $\sigma$: all $\tau$-open sets that do not contain $a$ are in $\sigma$ as well. And as $\operatorname{cl}_{\tau}(S \cap C)$ is a closed subset of the $\tau$-compact set $C$, it is also compact (under $\tau$, and hence under $\sigma$ as well, as it is coarser). Here we use the fact that $C$ (and its subsets) miss $a$, as we could assume from above.