# countably compact space

If $(X,\tau)$ is a topological space and $A \not\in \tau$, then $\tau(A)= \{U_1 \cup (U_2 \cap A ) : U_1, U_2 \in \tau \}$ is the topology on $X$ generated by the subbase ${A} \cup F$; $\tau(A)$ is called the simple extension of $\tau$ by $A$.

Can we say if $( X ,\tau)$ is countably compact and $A ‎\subseteq X$, then $( X, \tau(X - A ))$ will be countably compact ?

## 1 Answer

Not necessarily. Let $X=[0,1]$ with the usual topology, and let $A=X\setminus\{2^{-n}:n\in\omega\}$; then $\{2^{-n}:n\in\omega\}$ is an infinite, closed, discrete set in $\langle X,\tau(A)\rangle$.

• In " Maximal And minimal Topology " by " Douglas E.Cameron " noted that " A toplogical property $R$ has condition : ( 1 ) : if $(X,\tau)$ has property $R$ and $A \subset X$ has property $R$, Then $( X, \tau(X - A ))$ has property $R$. ( 2) : if a single point has property $R$. is countably compact true fo these conditons? – fatemeh Aug 30 '13 at 10:20
• @fatemeh: Yes. In fact, Cameron even says so. – Brian M. Scott Sep 1 '13 at 6:43