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A space $(X,\tau )$ is said to be minimal $KC$ if $(X,\tau)$ is $KC$ but no topology on $X$ which is strictly smaller than $\tau$ is $KC$.

Theorem : Every minimal KC space is compact.

Proof. Suppose for contradiction that $(X,\tau)$ is a minimal $KC$ space which is not compact. From Theorem $1$ below let $D$ be a discrete subset of $X$ with non-compact closure and $\mathcal{F}$ an ultrafilter such that $D \in \mathcal{F}$ and for every $C \in \mathcal{F}$, $C$ is not compact. $\mathcal{F}$ does not converge to any point in $D$, so let $x_{0} \in D$ and define a new topology $\sigma$ on $X$, such that: $$\sigma = \{U \in \tau:x_{0} \in U\in \mathcal{F}\} \cup \{U\in \tau: x_{0} \not\in U \}\;.$$ $(X,\sigma)$ is a $T_{1}$-topology which is strictly weaker than $\tau$. From Theorem $2$ below we have that every $\tau$-compact subset is also $\sigma$-closed. Finally, Theorem $3$ below says that there is no $\sigma$-compact subset which is not $\tau$-compact. And this proves that $\sigma$ is a $KC$ topology. It contradicts the hypothesis that $(X,\tau)$ is minimal $KC$.

Theorem $\bf 1$: Let $(X,\tau)$ be a $KC$ non-compact space. Then there is a discrete subset $D\subseteq X$, such that $D$ is not compact. Furthermore there is an ultrafilter $\mathcal{F}$ on $X$ such that $D\in\mathcal{F}$ and for every $C \in \mathcal{F}$, $C$ is not compact.

Theorem $\bf 2$: Let $(X,\tau)$ be a $KC$-space and $\mathcal{F}$ be an ultrafilter such that $\overline{F}_{\tau}$ is not compact for any $F \in \mathcal{F}$. Let $\sigma$ be a new topology on $X$ as above. If $K \subseteq X$ is $\tau$-compact, then it is $\sigma$-closed and the topologies $\tau$ and $\sigma$ agree on $K$.

Theorem $\bf 3$: Let $(X,\tau)$ be a minimal $KC$ space, $D$ a discrete subset with non-compact $\tau$-closure, and $\mathcal{F}$ an ultrafilter such that $D \in \mathcal{F}$ and $\mathcal{F}$ contains only sets with non-compact $\tau$-closures. Let $\sigma$ be a new topology on $X$ as above. Then every $\sigma$-compact subset is also $\tau$-compact.

  1. Why does $\mathcal F$ not converge to any point in $D$?

  2. Why in the last line "$\sigma$ is a $KC$ topology"?

Brian M. Scott
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fatemeh
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1 Answers1

3

  1. Let $x\in D$, and suppose that $\mathscr{F}$ converges to $x$; this means that $U\cap F\ne\varnothing$ whenever $U$ is an open nbhd of $x$ and $F\in\mathscr{F}$. $D$ is discrete, so there is a $U\in\tau$ such that $U\cap D=\{x\}$, and therefore $\{x\}=U\cap D\in\mathscr{F}$. But this is impossible, since $\{x\}$ is compact, and no member of $\mathscr{F}$ is compact. Thus, $\mathscr{F}$ does not converge to $x$.

  2. Suppose that $K\subseteq X$ is $\sigma$-compact; we want to show that $K$ is $\sigma$-closed. From Theorem $3$ we know that $K$ is $\tau$-compact, and then from Theorem $2$ we know that $K$ is $\sigma$-closed, which is exactly what we wanted. (You really should not have had to ask this question: everything that you needed is right there. No outside knowledge and no extra steps in the reasoning are required.)

Brian M. Scott
  • 631,399