Definition:Let $\tau$ be a topology on a set $X$, and let $\mathcal{B} \subset \tau$. Then $\mathcal{B}$ is a basis for the topology $\tau$ if and only if every open set in $\tau$ is the union of elements of $\mathcal{B}$. If $B \in \mathcal{B}$, we say $B$ is a basis element. Note that $B$ is an element of the basis of $\mathcal{B}$, but a subset of the set $X$.
Theorem(used in proof):
Let $(X,\tau)$ be a topological space, and let $\mathcal{B}$ be a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for $\tau$ if and only if
$(1) \mathcal{B} \subset \tau$ and
$(2)$ for each set $U$ in $\tau$ and point $p$ in $U$ there is a set $V$ in $\mathcal{B}$ such that $p \in V \subset U$.
Theorem(trying to prove):Suppose $X$ is a set and $\mathcal{B}$ is a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for some topology if and only if
$(1)$ each point of $X$ is in some element of $\mathcal{B}$ and
$(2)$ if $U$ and $V$ are sets in $\mathcal{B}$ and $p$ is a point in $U \cap V$, there is a set $W$ in $\mathcal{B}$ such that $p \in W \subset (U \cap V)$.
I am confident with the first part however the other direction of the biconditional has been confusing me.
($\Rightarrow$)
Since $\mathcal{B}$ is a basis, $X$ can be expressed as the union of elements of $\mathcal{B}$, so each point in $X$ is in some basis element. If $U,V \in\mathcal{B}$ since $U,V$ are open, $U \cap V$ is also open.Let $x \in U \cap V$. Since $U \cap V$ is open there exists a basis element $W$ such that $x \in W \subset U \cap V$.
($\Leftarrow$)
I have been struggling deciding what I need to show for this direction. Do I need to show for every point in every open set in $\tau$, there is a subset containing the point, contained in the open set, or do I need to show every element of $\tau$ can be expressed as a union of elements satisfying $(1)$ and $(2)$. Thinking about this, I realize it is wrong to already claim there is a set $\tau$, being a topology, without having proven it.
I will try and do the latter. Since each $x \in X$ is in a set satisfying $(1)$ and $(2)$, $X$ is the union of these sets. $\varnothing$ is in the union of these sets vacuously. I am having difficulty deciding what to do, to prove the second half of this theorem. Any hints will be appreciated.
Best Answer
You have a family $\mathscr{B}$ of subsets of $X$ such that each $x\in X$ belongs to at least one $B\in\mathscr{B}$, and whenever $p\in U\cap V$, where $U,V\in\mathscr{B}$, there is a $W\in\mathscr{B}$ such that $p\in W\subseteq U\cap V$. You want to prove that $\mathscr{B}$ is a base for a topology $\tau$ on $X$. Since this is the case if and only if
$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\,,\tag{1}$$
the natural approach is to define $\tau$ by $(1)$ and prove that $\tau$ is a topology on $X$. You’ve already observed that $\varnothing,X\in\tau$, so it remains to be shown that $U\cap V\in\tau$ whenever $U,V\in\tau$, and that $\bigcup\mathscr{U}\in\tau$ whenever $\mathscr{U}\subseteq\tau$.
It’s helpful to prove a small lemma first: if $B_0,B_1\in\mathscr{B}$, then there is a $\mathscr{U}\subseteq\mathscr{B}$ such that $B_0\cap B_1=\bigcup\mathscr{U}$ (and hence $B_0\cap B_1\in\tau$). I’ll leave this to you: it follows immediately from the second condition on $\mathscr{B}$.
Suppose that $U,V\in\tau$. Then there are $\mathscr{U},\mathscr{V}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$ and $V=\bigcup\mathscr{V}$. Then
$$\begin{align*} U\cap V&=\left(\bigcup\mathscr{U}\right)\cap\bigcup\mathscr{V}\\ &=\bigcup_{B\in\mathscr{U}}\left(B\cap\bigcup\mathscr{V}\right)\\ &=\bigcup_{B\in\mathscr{U}}\bigcup_{C\in\mathscr{V}}(B\cap C)\,.\tag{2} \end{align*}$$
By the lemma each of the sets $B\cap C$ in $(2)$ is the union of members of $\mathscr{B}$, so $U\cap V$ is a union of members of $\mathscr{B}$ and is therefore in $\tau$, as desired.
The proof that $\tau$ is closed under arbitrary unions is essentially the same as the last step of the previous argument.