# Topology question

• Jan 19th 2009, 05:07 PM
spoon737
Topology question
I'm trying to teach myself topology from Munkres' book (first edition, it's over 30 years old!) and it's going very slowly. I'm having some difficulty proving the following:

Let $\displaystyle \{T_\alpha\}$ be a collection of topologies on $\displaystyle X$. Show that there is a unique smallest topology containing all the collections $\displaystyle T_\alpha$, and a unique largest topology contained in all $\displaystyle T_\alpha$.

This is part b in a three part problem. The first part was to prove that if $\displaystyle \{T_\alpha\}$ is a collection of topologies on $\displaystyle X$ that $\displaystyle \bigcap T_\alpha$ is a topology on $\displaystyle X$, which I was able to do. So, imagine that result will be useful in what I'm trying to prove here.
• Jan 19th 2009, 07:56 PM
ThePerfectHacker
Quote:

Originally Posted by spoon737
I'm trying to teach myself topology from Munkres' book (first edition, it's over 30 years old!) and it's going very slowly. I'm having some difficulty proving the following:

Let $\displaystyle \{T_\alpha\}$ be a collection of topologies on $\displaystyle X$. Show that there is a unique smallest topology containing all the collections $\displaystyle T_\alpha$, and a unique largest topology contained in all $\displaystyle T_\alpha$.

This is part b in a three part problem. The first part was to prove that if $\displaystyle \{T_\alpha\}$ is a collection of topologies on $\displaystyle X$ that $\displaystyle \bigcap T_\alpha$ is a topology on $\displaystyle X$, which I was able to do. So, imagine that result will be useful in what I'm trying to prove here.

If $\displaystyle \{ T_{\alpha} \}$ is a collection of topologies on $\displaystyle X$ then $\displaystyle \cap T_{\alpha}$ is a topology on $\displaystyle X$. Notice that $\displaystyle \left( \cap T_{\alpha} \right) \subseteq T_{\beta}$ for each $\displaystyle \beta \in \alpha$ i.e. that this intersection of topologies is contained in each topology. Now if $\displaystyle T$ is a topology contained in each topology it means $\displaystyle T\subseteq T_{\beta}$ for each $\displaystyle \beta \in \alpha$ but then $\displaystyle T\subseteq \left( \cap T_{\alpha} \right)$. Thus, we see that any topology contained in all topologies must immediately be contained in the intersection of all topologies. Thus, $\displaystyle \cap T_{\alpha}$ is the largest topology contained in all topologies. From here uniqueness should quickly follow.

For the other problem realize that $\displaystyle \cup T_{\alpha}$ is not necessarily a topology! So we cannot use a similar around as we just did above except that $\displaystyle \cap$ is replaced by $\displaystyle \cup$. Instead, consider the discrete topology on $\displaystyle X$, and consider the intersection of all topologies containing $\displaystyle \cup T_{\alpha}$ (you need to show this is non-empty first). Once you establish this argue that the intersection of all these topologies is the smallest topology containing the collection of all topologies.
• Jan 19th 2009, 10:39 PM
spoon737
Oh wow, I can't believe I missed that first part. I had pretty much gotten to the end, but some how I missed the fact that $\displaystyle \cap T_{\alpha}$ is finer than any other topology contained in every $\displaystyle T_{\beta}$. It seems so obvious now!

Okay, for the second part, let's see if I get this.

Since the discrete topology contains every subset of $\displaystyle X$, it clearly contains $\displaystyle \cup T_{\alpha}$. So, the intersection of all topologies containing $\displaystyle \cup T_{\alpha}$, call it $\displaystyle I$, is non-empty. If $\displaystyle T$ is some topology containing $\displaystyle \cup T_{\alpha}$, then by definition, $\displaystyle I \subseteq T$. Thus, $\displaystyle I$ is the smallest topology containing all topologies in $\displaystyle \{T_\alpha\}$.

Does that look right?
• Jan 20th 2009, 09:53 AM
ThePerfectHacker
Quote:

Originally Posted by spoon737
Oh wow, I can't believe I missed that first part. I had pretty much gotten to the end, but some how I missed the fact that $\displaystyle \cap T_{\alpha}$ is finer than any other topology contained in every $\displaystyle T_{\beta}$. It seems so obvious now!

Okay, for the second part, let's see if I get this.

Since the discrete topology contains every subset of $\displaystyle X$, it clearly contains $\displaystyle \cup T_{\alpha}$. So, the intersection of all topologies containing $\displaystyle \cup T_{\alpha}$, call it $\displaystyle I$, is non-empty. If $\displaystyle T$ is some topology containing $\displaystyle \cup T_{\alpha}$, then by definition, $\displaystyle I \subseteq T$. Thus, $\displaystyle I$ is the smallest topology containing all topologies in $\displaystyle \{T_\alpha\}$.

Does that look right?

Yes. The reason why the intersection is non-empty is because $\displaystyle \mathcal{P}(X)$ is an example of a topology containing $\displaystyle \cup T_{\alpha}$.