Which of our four countability axioms does \(\mathbb{R}^\omega\) in the uniform topology satisfy?
Solution
Goal: Determine which of the four countability axioms (first-countable, second-countable, Lindelöf, separable) are satisfied by \(\mathbb{R}^\omega\) equipped with the uniform metric topology.
1. Metric characterization of the uniform topology:The uniform topology on \(\mathbb{R}^\omega\) is induced by the uniform metric:\[\bar{\rho}(\mathbf{x}, \mathbf{y}) = \sup_{n \in \mathbb{Z}_+} \bar{d}(x_n, y_n), \quad \text{where } \bar{d}(a, b) = \min\{|a - b|, 1\}.\]
2. \(\mathbb{R}^\omega\) is first-countable:Proof: Every metric space is first-countable. For each point \(\mathbf{x} \in \mathbb{R}^\omega\), the countable collection of open balls \(\{B_{\bar{\rho}}(\mathbf{x}, \frac{1}{k}) : k \in \mathbb{Z}_+\}\) forms a neighborhood basis at \(\mathbf{x}\).
3. Construction of an uncountable discrete subspace:Let \(S = \{0, 1\}^\omega \subset \mathbb{R}^\omega\) be the set of all binary sequences \(\mathbf{x} = (x_n)_{n=1}^\infty\) with \(x_n \in \{0, 1\}\).Proof:
3.1. The set \(S\) has cardinality \(|S| = 2^{\aleph_0} > \aleph_0\), so \(S\) is uncountable.
3.2. For any two distinct sequences \(\mathbf{x}, \mathbf{y} \in S\), there exists an index \(m \in \mathbb{Z}_+\) such that \(x_m \neq y_m\).
3.3. Then \(\bar{d}(x_m, y_m) = |x_m - y_m| = 1\), so \(\bar{\rho}(\mathbf{x}, \mathbf{y}) = \sup_{n} \bar{d}(x_n, y_n) = 1\).
3.4. The open balls \(\{B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2}) : \mathbf{x} \in S\}\) are pairwise disjoint in \(\mathbb{R}^\omega\).
4. \(\mathbb{R}^\omega\) is not second-countable:Proof: Let \(\mathcal{B}\) be any basis for the uniform topology on \(\mathbb{R}^\omega\). For each \(\mathbf{x} \in S\), the open set \(B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2})\) must contain a basis element \(U_\mathbf{x} \in \mathcal{B}\) with \(\mathbf{x} \in U_\mathbf{x} \subseteq B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2})\). Since the balls are pairwise disjoint (<1>3), the assignment \(\mathbf{x} \mapsto U_\mathbf{x}\) is injective. Hence \(|\mathcal{B}| \ge |S| = 2^{\aleph_0}\), so no countable basis exists.
5. \(\mathbb{R}^\omega\) is not separable:Proof: Let \(D \subset \mathbb{R}^\omega\) be any dense subset. For each \(\mathbf{x} \in S\), the non-empty open ball \(B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2})\) must contain a point \(d_\mathbf{x} \in D\). By pairwise disjointness, all \(d_\mathbf{x}\) are distinct, so \(|D| \ge |S| = 2^{\aleph_0}\). Thus no countable dense subset exists.
6. \(\mathbb{R}^\omega\) is not Lindelöf:Proof: For each \(\mathbf{x} \in S\), the set \(F_\mathbf{x} = \{\mathbf{x}\}\) is closed in \(\mathbb{R}^\omega\). Consider the open cover of \(\mathbb{R}^\omega\) given by \(\mathcal{U} = \{B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2}) : \mathbf{x} \in S\} \cup \{\mathbb{R}^\omega \setminus S\}\). Any subcover must contain every ball \(B_{\bar{\rho}}(\mathbf{x}, \frac{1}{2})\) because \(\mathbf{x} \in S\) belongs to no other member of \(\mathcal{U}\). Since \(S\) is uncountable, no countable subcover exists.
7. Conclusion:\(\mathbb{R}^\omega\) in the uniform topology satisfies the first countability axiom, but satisfies none of the other three (second-countable, separable, Lindelöf). Q.E.D.