1. ## Compactness II

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

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

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

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

I'd really appreciate some help!

2. Originally Posted by Showcase_22
$\displaystyle (X,T')$ compact gives:

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

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

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

3. Originally Posted by Showcase_22
$\displaystyle (X,T')$ compact gives:

Let $\displaystyle U$ be any open cover of $\displaystyle (X,T')$. Then there exists a finite subcover of $\displaystyle U$, say $\displaystyle U_1=\{u_1, \ldots, u_n\}$ such that $\displaystyle X \subset U_1 \subset \bigcup_{i=1}^n u_i$.
Are you sure this is what your notes say? The condition that X be compact is that there exist a finite subcover such that $\displaystyle X\subset \bigcup_{i=1}^n u_i$. There is no requirement that it be contained in one of the open sets so $\displaystyle X\subset u_1$ shouldn't be there.

Each $\displaystyle u_i$ 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 $\displaystyle u_i \in T'$.

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

4. Originally Posted by Showcase_22
$\displaystyle (X,T')$ compact gives:

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

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

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

I'd really appreciate some help!
Define $\displaystyle \varphiX,T')\to(X,T)$ by the identity map. Then clearly this is continuous. For, if $\displaystyle E\subseteq (X,T)$ is open then $\displaystyle E\in T\implies E=\varphi^{-1}(E)\in T'$ and thus $\displaystyle \varphi^{-1}(E)$ is open. Thus, we see that $\displaystyle (X,T)$ is the continuous image of a compact space, thus compact.