Let \(A\) be a closed subspace of \(X\). Show that if \(X\) is Lindelöf, then \(A\) is Lindelöf. Show by example that if \(X\) has a countable dense subset, \(A\) need not have a countable dense subset.
Solution
Goal: Prove that the Lindelöf property is inherited by closed subspaces, and provide a standard counterexample showing that separability (having a countable dense subset) is not hereditary to closed subspaces.
1. Closed subspaces of Lindelöf spaces are Lindelöf:Let \(X\) be Lindelöf and \(A \subseteq X\) be a closed subspace. Then \(A\) is Lindelöf.Proof:
1.1. Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in J}\) be an open cover of the subspace \(A\).
1.2. By definition of the subspace topology, for each \(\alpha \in J\) there exists an open set \(V_\alpha \subseteq X\) such that \(U_\alpha = V_\alpha \cap A\).
1.3. Since \(A \subseteq \bigcup_{\alpha \in J} V_\alpha\) and \(A\) is closed in \(X\), the complement \(X \setminus A\) is open in \(X\).
1.4. The family \(\mathcal{V} = \{V_\alpha\}_{\alpha \in J} \cup \{X \setminus A\}\) is an open cover of the entire space \(X\).
1.5. Because \(X\) is Lindelöf, there exists a countable subcover of \(X\):\[\mathcal{V}' = \{V_{\alpha_n}\}_{n=1}^\infty \cup \{X \setminus A\}.\]
1.6. Intersecting with \(A\), since \((X \setminus A) \cap A = \varnothing\):\[A = A \cap \left[ (X \setminus A) \cup \bigcup_{n=1}^\infty V_{\alpha_n} \right] = \bigcup_{n=1}^\infty (V_{\alpha_n} \cap A) = \bigcup_{n=1}^\infty U_{\alpha_n}.\]
1.7. Thus \(\{U_{\alpha_n}\}_{n=1}^\infty\) is a countable subcover of \(A\), proving \(A\) is Lindelöf.
2. Counterexample: Separability is not hereditary to closed subspaces.Proof:
2.1. Consider the Sorgenfrey plane \(X = \mathbb{R}_l \times \mathbb{R}_l\), where \(\mathbb{R}_l\) is the lower limit topology (Sorgenfrey line).
2.2. The countable subset \(D = \mathbb{Q} \times \mathbb{Q}\) is dense in \(X\) because every basic open rectangle \([a, b) \times [c, d)\) contains points of \(\mathbb{Q} \times \mathbb{Q}\). Thus \(X\) is separable.
2.3. Consider the “anti-diagonal” line \(L = \{(x, -x) \mid x \in \mathbb{R}\} \subset X\).
2.4. The subspace \(L\) is closed in \(X\): for any \((x, y) \notin L\), we have \(x + y \neq 0\); set \(\varepsilon = \frac{|x+y|}{2} > 0\), and the basic rectangle \([x, x+\varepsilon) \times [y, y+\varepsilon)\) is disjoint from \(L\), since any point \((u, -u) \in L\) in it would satisfy \(u \ge x\) and \(-u \ge y\), giving \(x + y \le u - u = 0\), contradicting \(x + y > 0\) (and symmetrically for \(x + y < 0\)).
2.5. The subspace topology induced on \(L\) is discrete: for any \(p = (x, -x) \in L\), the basic open set \(U = [x, x+1) \times [-x, -x+1)\) in \(X\) satisfies \(U \cap L = \{(x, -x)\} = \{p\}\).
2.6. In an uncountable discrete space, the only dense subset is the entire space itself, which is uncountable.
2.7. Therefore, the closed subspace \(L\) is not separable. Q.E.D.