Show that \([0, 1]\) is not limit point compact as a subspace of \(\mathbb{R}_\ell\).
Solution
Goal: Prove that the closed unit interval \([0, 1]\) endowed with the subspace topology from the lower limit line \(\mathbb{R}_\ell\) is not limit point compact.
1. Definition of limit point compactness:Proof:
1.1. A topological space \(X\) is limit point compact (also called countably compact for \(T_1\) spaces) if every infinite subset of \(X\) has at least one limit point in \(X\).
1.2. A point \(x \in X\) is a limit point of \(A \subseteq X\) if every open neighborhood \(U\) of \(x\) contains a point of \(A\) distinct from \(x\) (i.e. \((U \setminus \{x\}) \cap A \neq \emptyset\)).
2. Construction of an infinite subset \(A \subset [0, 1]\):Proof:
2.1. Define\[A = \left\{ 1 - \frac{1}{n} : n \in \mathbb{N}, \; n \ge 2 \right\} = \left\{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots \right\} \subset [0, 1].\]
2.2. Since the sequence \(x_n = 1 - 1/n\) is strictly increasing, \(A\) is an infinite subset of \([0, 1]\).
3. Verification that \(A\) has no limit points in \([0, 1]_{\mathbb{R}_\ell}\):Proof:We show that for every \(x \in [0, 1]\), there exists an open neighborhood \(U_x\) of \(x\) in \([0, 1]_{\mathbb{R}_\ell}\) such that \((U_x \setminus \{x\}) \cap A = \emptyset\).
3.1. Case \(x \in [0, 1/2)\): The set \(U_x = [x, 1/2) \cap [0, 1]\) is open in \([0, 1]_{\mathbb{R}_\ell}\), contains \(x\), and satisfies \(U_x \cap A = \emptyset\) because every element of \(A\) is \(\ge 1/2\).
3.2. Case \(x = 1 - 1/n \in A\) for some \(n \ge 2\): The set \(U_x = [1 - 1/n, 1 - 1/(n+1)) \cap [0, 1]\) is open in the lower limit subspace topology, contains \(x\), and contains no other element of \(A\), so \((U_x \setminus \{x\}) \cap A = \emptyset\).
3.3. Case \(x \in (1 - 1/n, 1 - 1/(n+1))\) for some \(n \ge 2\): The set \(U_x = [x, 1 - 1/(n+1)) \cap [0, 1]\) is open, contains \(x\), and satisfies \(U_x \cap A = \emptyset\).
3.4. Case \(x = 1\): The set \(U_1 = [1, 2) \cap [0, 1] = \{1\}\) is open in \([0, 1]_{\mathbb{R}_\ell}\) and contains no point of \(A\), so \((U_1 \setminus \{1\}) \cap A = \emptyset\).
4. Conclusion:Proof:The infinite subset \(A \subset [0, 1]\) has no limit point in \([0, 1]_{\mathbb{R}_\ell}\). Therefore \([0, 1]\) is not limit point compact in \(\mathbb{R}_\ell\).