Let \(X\) have a countable basis; let \(A\) be an uncountable subset of \(X\). Show that uncountably many points of \(A\) are limit points of \(A\).
Solution
Goal: Prove that if \(X\) is a second-countable topological space and \(A \subseteq X\) is uncountable, then uncountably many points of \(A\) are limit points of \(A\) (i.e. \(A \cap A'\) is uncountable).
1. Identification of non-limit points (isolated points of \(A\)):Proof:
1.1. Let \(I = A \setminus A'\) denote the set of points in \(A\) that are not limit points of \(A\).
1.2. By definition of limit points, for every \(x \in I\), there exists an open neighborhood \(U_x\) of \(x\) in \(X\) such that:\[U_x \cap A = \{x\}.\]
2. Proof that \(I\) is at most countable:Proof:
2.1. Let \(\mathcal{B} = \{B_n\}_{n=1}^\infty\) be a countable basis for the topology of \(X\).
2.2. For each \(x \in I\), choose a basis element \(B(x) \in \mathcal{B}\) such that \(x \in B(x) \subseteq U_x\).
2.3. Since \(x \in B(x) \subseteq U_x\), we have \(B(x) \cap A \subseteq U_x \cap A = \{x\}\). Because \(x \in B(x) \cap A\), it follows that:\[B(x) \cap A = \{x\}.\]
2.4. The assignment \(\phi: I \to \mathcal{B}\) defined by \(\phi(x) = B(x)\) is injective: if \(\phi(x) = \phi(y)\), then \(\{x\} = B(x) \cap A = B(y) \cap A = \{y\}\), forcing \(x = y\).
2.5. Because \(\mathcal{B}\) is countable and \(\phi\) is injective, the set \(I = A \setminus A'\) is at most countable.
3. Conclusion:
3.1. We partition \(A\) as:\[A = (A \cap A') \cup (A \setminus A') = (A \cap A') \cup I.\]
3.2. Since \(A\) is uncountable and \(I\) is countable, the set \(A \cap A'\) of points of \(A\) that are limit points of \(A\) must be uncountable. Q.E.D.