Compactness II

**Showcase_22** · March 13th 2010, 01:04 PM

Given topologies $T$ and $T'$ on a set $X$ with $T \subset T'$ , prove that if $(X,T')$ is compact then so is $(X,T)$ .

$(X,T')$ compact gives:

Let $U$ be any open cover of $(X,T')$ . Then there exists a finite subcover of $U$ , say $U_1=\{u_1, \ldots, u_n\}$ such that $X \subset U_1 \subset \bigcup_{i=1}^n u_i$ .

Each $u_i$ is open (is the "finite subcover" always open?) so each $u_i \in T'$ .

From here I get stuck. If we consider cases, if every $u_i \in T \subset T'$ then i'm done! However, if $u_i \in T'-T$ then i'm not really sure what to do. I don't even think this is possible since $u_i$ 's are open so should be in $T$ anyway.

I'd really appreciate some help!

**Opalg** · March 13th 2010, 01:32 PM

Originally Posted by Showcase_22

$(X,T')$ compact gives:

Let $U$ be any open cover of $(X,T')$ . Then there exists a finite subcover of $U$ , say $U_1=\{u_1, \ldots, u_n\}$ such that $X \subset U_1 \subset \bigcup_{i=1}^n u_i$ .

Each $u_i$ is open (is the "finite subcover" always open?) so each $u_i \in T'$ .

From here I get stuck. If we consider cases, if every $u_i \in T \subset T'$ then i'm done! However, if $u_i \in T'-T$ then i'm not really sure what to do. I don't even think this is possible since $u_i$ 's are open so should be in $T$ anyway.

I'd really appreciate some help!

I think you're trying to do this the wrong way round. You are told that (X,T') is compact, and you want to show that (X,T) is compact.

So let U be an open cover of (X,T). The members of U are T-open subsets of X. But $T\subset T'$ , so the members of U are T'-open. Therefore they form an open cover of (X,T'). So there is a finite subcover (and that's all that you need, to show that (X,T) is compact).

**HallsofIvy** · March 13th 2010, 01:33 PM

Originally Posted by Showcase_22

$(X,T')$ compact gives:

Let $U$ be any open cover of $(X,T')$ . Then there exists a finite subcover of $U$ , say $U_1=\{u_1, \ldots, u_n\}$ such that $X \subset U_1 \subset \bigcup_{i=1}^n u_i$ .

Are you sure this is what your notes say? The condition that X be compact is that there exist a finite subcover such that $X\subset \bigcup_{i=1}^n u_i$ . There is no requirement that it be contained in one of the open sets so $X\subset u_1$ shouldn't be there.

Each $u_i$ is open (is the "finite subcover" always open?)

You are misunderstanding- an "open cover" is a collection of sets in the topology, not a set in the topology itself so the word "open" does not apply to U itself but to the sets in U. Since U is an open cover, every set in it is open so every set in any sub-cover of U is also open.

so each $u_i \in T'$ .

From here I get stuck. If we consider cases, if every $u_i \in T \subset T'$ then i'm done! However, if $u_i \in T'-T$ then i'm not really sure what to do. I don't even think this is possible since $u_i$ 's are open so should be in $T$ anyway.

I'd really appreciate some help!

What exactly are you trying to prove? If you want to prove "if X is compact in T' then it is compact in $T\subset T'$ you have it the wrong way around. Given that there is a finite subcover of X in T', it does NOT follow that there is a finite subcover in T. There might be some open set in that subcover in T' that is not in T.

It is true that if X is compact in T' and $T'\subset T$ then X is compact in T.

**Drexel28** · March 13th 2010, 09:41 PM

Originally Posted by Showcase_22

$(X,T')$ compact gives:

Let $U$ be any open cover of $(X,T')$ . Then there exists a finite subcover of $U$ , say $U_1=\{u_1, \ldots, u_n\}$ such that $X \subset U_1 \subset \bigcup_{i=1}^n u_i$ .

Each $u_i$ is open (is the "finite subcover" always open?) so each $u_i \in T'$ .

From here I get stuck. If we consider cases, if every $u_i \in T \subset T'$ then i'm done! However, if $u_i \in T'-T$ then i'm not really sure what to do. I don't even think this is possible since $u_i$ 's are open so should be in $T$ anyway.

I'd really appreciate some help!

Define $\varphi<img src=$ X,T')\to(X,T)" alt="\varphi

X,T')\to(X,T)" /> by the identity map. Then clearly this is continuous. For, if $E\subseteq (X,T)$ is open then $E\in T\implies E=\varphi^{-1}(E)\in T'$ and thus $\varphi^{-1}(E)$ is open. Thus, we see that $(X,T)$ is the continuous image of a compact space, thus compact.

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