Countably Compact Sets

• Aug 27th 2013, 01:26 PM
Aryth
Countably Compact Sets
I'm working on this problem, and I'm not sure how to solve it.

Let $(X,\Omega)$ be a topological space and let $A \subseteq X$ be countably compact. Prove that every subset of $A$ that is closed relative to $\Omega$ is also countably compact.

I have a similar proof where this problem is stated without the 'countably' part. I'm unsure how to proceed with countably compact sets... Any help would be appreciated.
• Aug 27th 2013, 01:47 PM
Plato
Re: Countably Compact Sets
Quote:

Originally Posted by Aryth
Let $(X,\Omega)$ be a topological space and let $A \subseteq X$ be countably compact. Prove that every subset of $A$ that is closed relative to $\Omega$ is also countably compact.

As always this depends upon the definitions in use.
If $B\subset A$ that is closed relative to $\Omega$. Can you show that any countable open cover of $B$ has a finite sub cover all relative to $\Omega~?$
• Aug 27th 2013, 05:44 PM
Aryth
Re: Countably Compact Sets
Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

Suppose that $A \subseteq X$ is countably compact and let $B \subset A$ be closed. We will let $\mathcal{F} = \{F_i : i \in \mathbb{N}\}$ be a countable cover of $A$. Since $X \backslash B$ is open, it follows that $\mathcal{F}$ along with $X \backslash B$ is a countable open cover of $A$. We know that $A$ is countably compact, so $A$ can be covered by a finite number of sets from $\mathcal{F}$, $F_1, F_2, \cdots , F_n$, and possibly $X \backslash B$. Since $B \subset A$, $F_1, F_2, \cdots, F_n$ covers $B$. Therefore, $B$ is countably compact.
• Aug 27th 2013, 07:00 PM
Plato
Re: Countably Compact Sets
Quote:

Originally Posted by Aryth
Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

Suppose that $A \subseteq X$ is countably compact and let $B \subset A$ be closed. We will let $\mathcal{F} = \{F_i : i \in \mathbb{N}\}$ be a countable cover of $A$. Since $X \backslash B$ is open, it follows that $\mathcal{F}$ along with $X \backslash B$ is a countable open cover of $A$. We know that $A$ is countably compact, so $A$ can be covered by a finite number of sets from $\mathcal{F}$, $F_1, F_2, \cdots , F_n$, and possibly $X \backslash B$. Since $B \subset A$, $F_1, F_2, \cdots, F_n$ covers $B$. Therefore, $B$ is countably compact.

Good. That will work.