Show that every second countable space is first countable.
The original exercise asked for the converse, which is false: an uncountable discrete space is first countable but not second countable.
Solution
Goal: Prove that every second-countable space \(X\) is first-countable.
1. Setting and hypothesis:Let \(X\) be a second-countable topological space.By definition, \(X\) possesses a countable basis \(\mathcal{B} = \{B_n\}_{n=1}^\infty\) for its topology.
2. Construction of a countable local basis at an arbitrary point:Proof:
2.1. Let \(x \in X\) be an arbitrary point.
2.2. Define the subcollection of basic open sets containing \(x\):\[\mathcal{B}_x = \{B \in \mathcal{B} \mid x \in B\}.\]
2.3. Because \(\mathcal{B}_x \subseteq \mathcal{B}\) and \(\mathcal{B}\) is countable, \(\mathcal{B}_x\) is a countable collection of open neighborhoods containing \(x\).
3. Verification of the local basis property:Proof:
3.1. Let \(U \subseteq X\) be an arbitrary open neighborhood of \(x\).
3.2. Since \(\mathcal{B}\) is a topological basis for \(X\), there exists a basic open set \(B \in \mathcal{B}\) such that \(x \in B \subseteq U\).
3.3. Because \(x \in B\), \(B\) belongs to the subcollection \(\mathcal{B}_x\) by construction.
3.4. Thus, every open neighborhood \(U\) of \(x\) contains an element of \(\mathcal{B}_x\).
4. Conclusion:\(\mathcal{B}_x\) is a countable local basis at \(x\). Since \(x \in X\) was arbitrary, \(X\) is first-countable. Q.E.D.