Base for topologies? Theorem Let $\mathcal{B}$ be a class of subsets of a nonempty set $X.$ Then $\mathcal{B}$ will be a base for some topology on $X$ if and only if:
$i.)$ $X = \displaystyle\bigcup_{\alpha \in I} B_\alpha$
$ii.)$ For any $B_1, B_2 \in \mathcal{B},$ then $B_1 \cap B_2$ is the union of members of $\mathcal{B}$.

I came across this theorem in the General Topology book from Schaum's. It contains no proof and I'm interested in proving it. But there's a little ambiguity about this theorem.
Suppose $i.)$ and $ii.)$ are true. Is this theorem saying there exists a topology that satisfies this collection $\mathcal{B}$ as its base or are we saying that we know there is a topology and have to prove $\mathcal{B}$ exists.
If what I'm saying doesn't make sense, let me reword my question: are we assuming $\mathcal{B}$ exists and proving $\tau$ is a topology on $X$? Or are we assuming $\tau$ is a topology and showing that $\mathcal{B}$ is its base?
 A: The statement is of the form

$\mathcal{B}$ will be a base for some topology on $X$ if and only if i and ii.

There are two directions to prove.  You as about the direction where we assume the truth of i and ii.  Then the conclusion should be that there exists a topology for which $\mathcal{B}$ is a base.
Notice that $\mathcal{B}$ is introduced and exists before the biimplication.  Consequently, the complete context for the proof in the direction you ask about is $\mathcal{B}$ is a class of subsets of a nonempty set $X$, i is true, and ii is true.  You do not have to prove $\mathcal{B}$ exists because you already have it in your hand.
To your rewording, in the direction you ask about, you have $\mathcal{B}$, the truth of i, and the truth of ii.  You wish to show the existence of a topology of $X$, which you have called $\tau$, having $\mathcal{B}$ as its base.
A: The theorem asserts an equivalence of two things (‘if and only if’), so in fact it is saying both, depending on which direction of the equivalence you’re considering.
Note that if $\tau$ is a topology on $X$, then $\tau\setminus\{\varnothing\}$ is a family of non-empty subsets of $X$ satisfying both conditions, and clearly
$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\tau\setminus\{\varnothing\}\right\}\,,$$
so $\tau\setminus\{\varnothing\}$ is a base for $\tau$. This shows half of the theorem: that every topology on a set $X$ has a base of non-empty sets satisfying both conditions.
The other (and more interesting) half of the theorem says that if $\mathscr{B}$ is a family of non-empty subsets of $X$ that satisfies both conditions, then there is a topology $\tau$ on $X$ such that $\mathscr{B}$ is a base for $\tau$. Of course if $\tau$ is a topology on $X$, saying that $\mathscr{B}$ is a base for $\tau$ is simply saying that
$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\,,$$
so this direction of the theorem is actually saying that if $\mathscr{B}$ is a family of non-empty subsets of $X$ that satisfies both conditions, then
$$\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}$$
is a topology on $X$. The conditions are precisely what is needed to ensure that this is the case.
