Equivalent Definition of Basis of a Topology

Question

If we accept the definition of basis for the topology $\tau$ as stated in the picture, how can we prove Theorem 3.1?Here is my approach:

($\Rightarrow$)The first condition holds by definition. Let $U$ be open in $\tau$ such that $p\in U$ and $V\in\beta$ such that $p\in V$.(Right here, Am I allowed to make such a supposition? Is there any need to utilize axiom of choice? If yes, how should it be applied?) Since by definition $U=\cup_{B\in\beta}B$ and $V\in\beta$, then $p\in V\subset U$. But then I am lost on how to prove the second part. Could you please help me in completing the proof and correct my incomplete proof if necessary?

One small note: Instead of Greek letters, you can use the command \mathcal. For example, \mathcal{T} (within math mode) gives you $\mathcal{T}$. Similarly, \mathcal{B} gives $\mathcal{B}$. — Aryaman Maithani, Oct 09 '21 at 14:54
You apply the definition of union, not the axiom of choice…. — Henno Brandsma, Oct 09 '21 at 15:10

score 0 · Answer 1 · answered Oct 09 '21 at 14:47

By assumption, $U=\bigcup_iB_i$ for some basic open sets $B_i\in\mathfrak B$. Then, by definition of union, for every $x\in U$ there must be an index $i$ such that $x\in B_i$, and the union also implies $B_i\subseteq U$.

For the converse, we indeed use the axiom of choice: for each $x\in U$ choose a set $V_x\in\mathfrak B$ with the given property: $x\in V_x\subseteq U$.
Then verify that $U=\bigcup_{x\in U}V_x$.

Equivalent Definition of Basis of a Topology

1 Answers1