Recall that \(\mathbb{R}_K\) denotes \(\mathbb{R}\) in the \(K\)-topology.
Show that \([0, 1]\) is not compact as a subspace of \(\mathbb{R}_K\).
Show that \(\mathbb{R}_K\) is connected. [Hint: \((-\infty, 0)\) and \((0, \infty)\) inherit their usual topologies as subspaces of \(\mathbb{R}_K\).]
Show that \(\mathbb{R}_K\) is not path connected.
Solution
Goal: Analyze compactness of \([0, 1]\), connectedness of \(\mathbb{R}_K\), and non-path-connectedness of \(\mathbb{R}_K\).
1. Part (a): \([0, 1]\) is not compact in \(\mathbb{R}_K\).Proof:
1.1. The basis for \(\mathbb{R}_K\) includes sets of the form \((a, b) \setminus K\), where \(K = \{\frac{1}{n} : n \in \mathbb{Z}_+\}\).
1.2. Define the open cover \(\mathcal{U}\) of \([0, 1]\) by:\[\mathcal{U} = \{(-1, 1) \setminus K\} \cup \left\{\left(\frac{1}{n+1}, \frac{1}{n-1}\right) : n \ge 2\right\} \cup \left\{\left(\frac{1}{2}, 2\right)\right\}.\]
1.3. \(\mathcal{U}\) is an open cover: \((-1, 1) \setminus K\) contains \(0\) and all points of \([0, 1] \setminus K\); each point \(\frac{1}{n} \in K\) with \(n \ge 2\) belongs to \((\frac{1}{n+1}, \frac{1}{n-1})\); and \(1 \in (\frac{1}{2}, 2)\).
1.4. Any subcover must contain \((-1, 1) \setminus K\) to cover \(0\).
1.5. Since \((-1, 1) \setminus K\) contains no point of \(K\), and each interval \((\frac{1}{n+1}, \frac{1}{n-1})\) contains only the single point \(\frac{1}{n} \in K\), any subcover must contain every interval \((\frac{1}{n+1}, \frac{1}{n-1})\) for all \(n \ge 2\).
1.6. Thus no finite subcover exists, so \([0, 1]\) is not compact in \(\mathbb{R}_K\).
2. Part (b): \(\mathbb{R}_K\) is connected.Proof:
2.1. The subspace topology on \((-\infty, 0)\) is the standard Euclidean topology since \((-\infty, 0) \cap K = \emptyset\), so \((-\infty, 0)\) is connected.
2.2. The subspace topology on \((0, \infty)\) is also the standard Euclidean topology because \(K\) has no limit points in \((0, \infty)\), so \((0, \infty)\) is connected.
2.3. Suppose \(\mathbb{R}_K = U \cup V\) is a separation of \(\mathbb{R}_K\) into disjoint open sets. Without loss of generality, let \(0 \in U\).
2.4. Since \(0 \in U\) and \(U\) is open, \(U\) contains a basic neighborhood \((-\varepsilon, \varepsilon) \setminus K\) for some \(\varepsilon > 0\).
2.5. Then \(U \cap (-\infty, 0) \supseteq (-\varepsilon, 0) \neq \emptyset\). Since \((-\infty, 0)\) is connected, \((-\infty, 0) \subseteq U\).
2.6. Similarly, \(U \cap (0, \infty) \supseteq ((0, \varepsilon) \setminus K) \neq \emptyset\). Since \((0, \infty)\) is connected, \((0, \infty) \subseteq U\).
2.7. Thus \(\mathbb{R} = (-\infty, 0) \cup \{0\} \cup (0, \infty) \subseteq U\), so \(V = \emptyset\).
2.8. Hence no separation exists, so \(\mathbb{R}_K\) is connected.
3. Part (c): \(\mathbb{R}_K\) is not path connected.Proof:
3.1. Suppose there exists a continuous path \(\gamma: [0, 1] \to \mathbb{R}_K\) with \(\gamma(0) = 0\) and \(\gamma(1) = 1\).
3.2. Since the topology of \(\mathbb{R}_K\) is strictly finer than the standard topology on \(\mathbb{R}\), \(\gamma\) is also continuous as a map into the standard Euclidean space \(\mathbb{R}\).
3.3. By the Intermediate Value Theorem, the image \(\gamma([0, 1])\) is a connected subset of standard \(\mathbb{R}\) containing \(0\) and \(1\), so \([0, 1] \subseteq \gamma([0, 1])\).
3.4. Since \([0, 1]\) is compact in standard \(\mathbb{R}\) and \(\gamma: [0, 1] \to \mathbb{R}_K\) is continuous, the image \(\gamma([0, 1])\) is compact in \(\mathbb{R}_K\).
3.5. The interval \([0, 1]\) is closed in \(\mathbb{R}_K\) because \(\mathbb{R}_K \setminus [0, 1] = (-\infty, 0) \cup (1, \infty)\) is open in \(\mathbb{R}_K\).
3.6. As a closed subset of the compact space \(\gamma([0, 1])\), \([0, 1]\) must be compact in \(\mathbb{R}_K\).
3.7. This directly contradicts
1.
1.8. Thus no continuous path connects \(0\) to \(1\), so \(\mathbb{R}_K\) is not path connected. Q.E.D.