Find a nondiscrete space that has a countably locally finite basis but does not have a countable basis.
Solution
Let \(X\) be the set \(\RR\) with the topology in which \(U \subseteq X\) is open if and only if \(0 \notin U\) or \((-\varepsilon, \varepsilon) \subseteq U\) for some \(\varepsilon > 0\). Every point \(x \neq 0\) is isolated, and the intervals \((-\varepsilon, \varepsilon)\) form a neighborhood basis at \(0\).
1. \(X\) is not discrete.
Every open set containing \(0\) contains an interval \((-\varepsilon, \varepsilon)\), so \(\{0\}\) is not open.
2. For \(n \ge 1\), the collection \(\mathcal{S}_n = \{\{x\} : \abs{x} \ge 1/n\}\) is locally finite.
A point \(x \neq 0\) has the neighborhood \(\{x\}\), which meets only the member \(\{x\}\) of \(\mathcal{S}_n\), if any. The point \(0\) has the neighborhood \((-1/n, 1/n)\), which meets no member of \(\mathcal{S}_n\).
3. \(\mathcal{B} = \bigcup_{n \ge 1} \mathcal{S}_n \cup \{(-1/n, 1/n) : n \ge 1\}\) is a countably locally finite basis for \(X\).
Every \(x \neq 0\) satisfies \(\abs{x} \ge 1/n\) for some \(n\), so \(\{x\} \in \mathcal{B}\); the intervals \((-1/n, 1/n)\) form a neighborhood basis at \(0\). Hence \(\mathcal{B}\) is a basis. It is the union of the collections \(\mathcal{S}_n\), which are locally finite by step 2, and the collections \(\{(-1/n, 1/n)\}\), each of which has one member and is locally finite.
4. \(X\) has no countable basis.
For each \(x \neq 0\), the set \(\{x\}\) is open, so every basis contains a member \(B\) with \(x \in B \subseteq \{x\}\), that is, \(B = \{x\}\). There are uncountably many such \(x\).
5. Q.E.D.
By steps 1, 3, and 4, \(X\) is a nondiscrete space with a countably locally finite basis and no countable basis.