$\begingroup$

Theorem Any collection of subsets $\mathcal{A}$ of a nonempty set $X$ forms the subbasis for a unique topology $\tau$ on $X$.

This theorem is absolutely amazing to me. I really enjoy the idea of it as a powerful tool, but I have come up with a counterexample that I just can't get over.

So the theorem states that any collection of subsets of a nonempty $X$ form a subbasis for a unique topology on $X$. The emphasis there is any. So, consider the following counterexample:

Let $X= \{a,b,c,d,e\}$ and let $\mathcal{A}=\{\{a\}\}$. Clearly, this is a collection of subsets of $X$. Assume that, by our theorem, then $\mathcal{A}=\{\{a\}\}$ is a subbasis for some topology on $X$.

Okay, so since $\mathcal{A}$ is a subbasis of some topology on $X$, let's try taking intersections of members of $\mathcal{A}$.

Well, $\{a\}\cap\{a\}=\{a\}$.

Then our basis for our topology is $\mathcal{B} = \{\{a\}\}$

This is problematic because this means that our basis $\mathcal{B}$ is just $\{a\}$, but note that $\displaystyle\bigcup \mathcal{B} = \{a\}$ and $\{a\} \neq X.$ Therefore, $X \not \in \tau.$

How do we get $X$ in $\tau$? Is my counterexample logically consistent?