Show that if \(X\) is an infinite wedge of circles, then \(X\) does not satisfy the first countability axiom.
Solution
Goal: Prove that an infinite wedge sum of circles \(X = \bigvee_{\alpha \in J} S_\alpha^1\) fails to be first-countable at the common wedge basepoint \(p\).
1. Quotient topology on the wedge sum:Proof:
1.1. By definition of the wedge sum \(X = (\coprod_{\alpha \in J} S_\alpha^1) / \{p_\alpha\}\), a subset \(U \subseteq X\) containing the basepoint \(p\) is open in \(X\) if and only if for every \(\alpha \in J\), the intersection \(U \cap S_\alpha^1\) is open in \(S_\alpha^1\).
2. Setting up proof by contradiction:Proof:
2.1. Suppose for contradiction that \(p\) has a countable local neighborhood basis \(\{U_n\}_{n=1}^\infty\) of open sets in \(X\).
2.2. Since \(J\) is infinite, choose a countably infinite subset of distinct indices \(\{\alpha_n\}_{n=1}^\infty \subseteq J\).
3. Diagonal construction of an open neighborhood \(V\):Proof:
3.1. For each \(n \ge 1\), \(U_n \cap S_{\alpha_n}^1\) is an open neighborhood of \(p\) in \(S_{\alpha_n}^1\).
3.2. Since \(S_{\alpha_n}^1 \cong S^1\) has no isolated points, choose a point \(x_n \in (U_n \cap S_{\alpha_n}^1) \setminus \{p\}\).
3.3. Define \(V \subseteq X\) by setting its intersection with each circle:\[V \cap S_{\alpha_n}^1 = S_{\alpha_n}^1 \setminus \{x_n\} \quad \text{for all } n \ge 1,\]\[V \cap S_\beta^1 = S_\beta^1 \quad \text{for all } \beta \in J \setminus \{\alpha_n \mid n \ge 1\}.\]
3.4. For every \(\alpha \in J\), \(V \cap S_\alpha^1\) is open in \(S_\alpha^1\) and contains \(p\) (since \(x_n \neq p\)).
3.5. By the definition of the quotient topology in 1, \(V\) is an open neighborhood of \(p\) in \(X\).
4. Deriving the contradiction:Proof:
4.1. If \(\{U_n\}_{n=1}^\infty\) were a local neighborhood basis at \(p\), there would exist an index \(k \ge 1\) such that \(U_k \subseteq V\).
4.2. By construction, \(x_k \in U_k\).
4.3. However, \(x_k \notin S_{\alpha_k}^1 \setminus \{x_k\} = V \cap S_{\alpha_k}^1\), so \(x_k \notin V\).
4.4. This contradicts \(U_k \subseteq V\).
5. Conclusion:\(X\) has no countable local basis at \(p\), and therefore \(X\) is not first-countable. Q.E.D.