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?