Many spaces have countable bases; but no \(T_1\) space has a locally finite basis unless it is discrete. Prove this fact.
Solution
Let \(X\) be a \(T_1\) space with a locally finite basis \(\mathcal B\), and fix \(x \in X\).
1. Only finitely many members \(B_1, \ldots, B_k\) of \(\mathcal B\) contain \(x\), and \(k \ge 1\).
By local finiteness, some neighborhood \(U\) of \(x\) meets only finitely many members of \(\mathcal B\). Every member of \(\mathcal B\) that contains \(x\) meets \(U\). Since \(\mathcal B\) is a basis and \(X\) is open, some member of \(\mathcal B\) contains \(x\).
2. For each \(y \neq x\), some \(B_i\) does not contain \(y\).
Since \(X\) is \(T_1\), \(\{y\}\) is closed, so \(X \setminus \{y\}\) is an open set containing \(x\). It contains a member \(B\) of \(\mathcal B\) with \(x \in B\), and \(B\) is one of \(B_1, \ldots, B_k\) by step 1.
3. \(\{x\} = B_1 \cap \cdots \cap B_k\), so \(\{x\}\) is open.
Each \(B_i\) contains \(x\), and by step 2 no \(y \neq x\) lies in every \(B_i\). A finite intersection of open sets is open.
4. Q.E.D.
By step 3, every singleton of \(X\) is open, so \(X\) is discrete.