Any collection of subsets of a set is a subbasis for a topology 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?
 A: The whole space has to be included by force, or by hand.  Or the empty set and whole set can be interpreted as the vacuous intersection and union.
The point is that any subset of the power set generates a unique topology:   the intersection of all topologies containing it, or the smallest such topology.
A: The theorem is only valid if you accept that $\bigcap \varnothing = X$. Otherwise, the theorem should be stated as:

Any collection of subsets $\mathcal A$ of a non-empty set $X$ such that $\bigcup \mathcal A = X$  forms the subbasis for a unique topology on $X$.

In this case, the collection $\mathcal B$ of all finite intersections of elements in $\mathcal A$ (formally, $\mathcal B : = \{\bigcap F : F \textrm{ is a subset of $\mathcal A$ with } 1 \leq |F| < \aleph_0\}$) satisfies the conditions of being a basis for a topology on $X$:

*

*Given $x \in X$, the condition $\bigcup \mathcal A = X$ says that there exists $A \in \mathcal A$ with $x \in A$, but $A = \bigcap \{A\} \in \mathcal B$.

*If $B_1$ and $B_2$ are in $\mathcal B$, then $B_1 \cap B_2 \in \mathcal B$ as you can check.

