Give \(\mathbb{R}^\omega\) the box topology. Let \(\mathbb{Q}^\infty\) denote the subspace consisting of sequences of rationals that end in an infinite string of 0’s. Which of our four countability axioms does this space satisfy?
Solution
1. The space \(\mathbb{Q}^\infty\) is countable.
1.1. \(\mathbb{Q}^\infty = \bigcup_{k=1}^\infty X_k\), where \(X_k = \{(x_n) \in \mathbb{Q}^\omega : x_n = 0 \text{ for all } n > k\}\).
definition of sequences ending in infinite strings of zeros.
1.2. For each \(k\), \(X_k \cong \mathbb{Q}^k\) is countable, being a finite product of countable sets.
countability of \(\mathbb{Q}\) and finite cartesian products of countable sets.
1.3. \(\mathbb{Q}^\infty\) is a countable union of countable sets, hence countable.
1 and 2.
2. \(\mathbb{Q}^\infty\) is separable.
\(\mathbb{Q}^\infty\) is a countable subset of itself and is dense in itself, so \(\mathbb{Q}^\infty\) has a countable dense subset.
3. \(\mathbb{Q}^\infty\) is Lindelöf.
3.1. Let \(\mathcal{U}\) be an open cover of \(\mathbb{Q}^\infty\).
setup.
3.2. For each point \(x \in \mathbb{Q}^\infty\), choose an open set \(U_x \in \mathcal{U}\) containing \(x\).
\(\mathcal{U}\) covers \(\mathbb{Q}^\infty\).
3.3. The subcollection \(\{U_x : x \in \mathbb{Q}^\infty\}\) covers \(\mathbb{Q}^\infty\) and is countable since \(\mathbb{Q}^\infty\) is countable by
1.
image of a countable set under a choice function.
1.4. Hence every open cover has a countable subcover.
3.
4. \(\mathbb{Q}^\infty\) is not first-countable (and hence not second-countable).
4.1. A basic open neighborhood of \(\mathbf{0} = (0, 0, \dots)\) in the subspace box topology has the form \(V = \left(\prod_{n=1}^\infty (-\varepsilon_n, \varepsilon_n)\right) \cap \mathbb{Q}^\infty\) for positive reals \(\varepsilon_n > 0\).
definition of the box topology and subspace topology.
4.2. Suppose \(\{B_k\}_{k=1}^\infty\) is a countable neighborhood basis at \(\mathbf{0}\).
hypothesis for contradiction.
4.3. For each \(k \ge 1\), choose a basic neighborhood \(V_k = \left(\prod_{n=1}^\infty (-\varepsilon_{k,n}, \varepsilon_{k,n})\right) \cap \mathbb{Q}^\infty \subseteq B_k\) with \(\varepsilon_{k,n} > 0\).
\(\{B_k\}\) are neighborhoods of \(\mathbf{0}\).
4.4. Define \(\delta_n = \frac{1}{2} \varepsilon_{n,n} > 0\) for each \(n \ge 1\), and let \(W = \left(\prod_{n=1}^\infty (-\delta_n, \delta_n)\right) \cap \mathbb{Q}^\infty\).
\(W\) is an open neighborhood of \(\mathbf{0}\) in \(\mathbb{Q}^\infty\).
4.5. For each \(k \ge 1\), choose \(q_k \in \mathbb{Q}\) such that \(\delta_k < |q_k| < \varepsilon_{k,k}\).
\(\delta_k = \frac{1}{2}\varepsilon_{k,k} < \varepsilon_{k,k}\) and \(\mathbb{Q}\) is dense in \(\mathbb{R}\).
4.6. The point \(y^{(k)} = (0, \dots, 0, q_k, 0, \dots)\) (with \(q_k\) in the \(k\)-th coordinate) lies in \(\mathbb{Q}^\infty\) and satisfies \(y^{(k)} \in V_k \subseteq B_k\).
\(|y^{(k)}_n| = 0 < \varepsilon_{k,n}\) for \(n \neq k\) and \(|y^{(k)}_k| = |q_k| < \varepsilon_{k,k}\).
4.7. But \(|y^{(k)}_k| = |q_k| > \delta_k\), so \(y^{(k)} \notin W\).
definition of \(W\).
4.8. Thus \(B_k \not\subseteq W\) for all \(k \ge 1\), contradicting that \(\{B_k\}\) is a neighborhood basis at \(\mathbf{0}\).
4.7.
4.9. Hence \(\mathbb{Q}^\infty\) is not first-countable, and therefore not second-countable.
8 and second-countability implies first-countability.
5. Conclusion: \(\mathbb{Q}^\infty\) satisfies the Lindelöf and separability axioms, but does not satisfy the first-countability or second-countability axioms.
2, 3, 4.
Q.E.D.