J. Munkres. Topology (2013) Chapter 2: Topological Spaces and Continuous Functions. 14: The Order Topology

Definition

Let \(X\) be a linearly ordered set with more than one element. The order topology on \(X\) is the topology generated by the basis consisting of:

  1. All open intervals
  2. All intervals of the form \([\bot, b)\) where \(\bot\) is the smallest element of \(X\), if this exists.
  3. All intervals of the form \((a, \top]\) where \(\top\) is the greatest element of \(X\), if this exists.

Example 1

The standard topology on \(\mathbb{R}\) is the order topology.

The order topology is the topology generated by the subbasis of all open rays.

Comments

Post a Comment

Popular posts from this blog

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 32: Normal Spaces

Theorem 32.1 Every second countable regular space is normal. Proof Let \(X\) be a second countable regular space. Pick a countable basis \(\mathcal{B}\). Let \(A\) and \(B\) be disjoint closed sets in \(X\). Pick a countable covering \(\{U_1, U_2, \ldots\}\) of \(A\) by open sets whose closures are disjoint from \(B\). Let \(\{U_1, U_2, \ldots\} = \{U \in \mathcal{B} \mid \overline{U} \cap B = \emptyset\}\). Prove: this covers \(A\). Let \(x \in A\). Prove: there exists \(U \in \mathcal{B}\) such that \(x \in U\) and \(\overline{U} \cap B = \emptyset\). Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(B \subseteq V\). By regularity. Pick \(U' \in \mathcal{B}\) such that \(x \in U' \subseteq U\). \(x \in U'\) and \(\overline{U'} \cap B \subseteq \overline{U} \cap B \subseteq (X - V) \cap B = \emptyset\) Pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets whose ...
Read more

J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 27: Compact Subspaces of the Real Line

Theorem 27.1 Let \(X\) be a linearly ordered set with the least upper bound property in the order topology. Every closed interval in \(X\) is compact. Proof Let \(a,b \in X\) with \(a Let \(\mathcal{A}\) be a covering of \([a,b]\) by sets open in \(X\). \( \langle 1 \rangle 1 \) For any \(x \in [a,b)\), there exists \(y \in (x,b]\) such that \([x,y]\) can be covered by at most two elements of \(\mathcal{A}\). Let \(x \in [a,b)\). Pick \(U \in \mathcal{A}\) such that \(x \in U\). Pick \(y > x\) such that \([x,y) \subseteq U\). Assume w.l.o.g. \(y \leq b\). Pick \(V \in \mathcal{A}\) such that \(y \in V\). \([x,y]\) is covered by \(U\) and \(V\). Let \(C = \{y \in (a,b] \mid [a,y] \text{ can be covered by finitely many elements of } \mathcal{A} \}\). Let \(c = \sup C\). \(C\) is nonempty. Pick \(y \in (a,b]\) such that \([a,y]\) can be covered by at most two elements of \(\mathcal{A}\). \(\...
Read more

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 31: The Separation Axioms

Definition A topological space \(X\) is regular iff it is \(T_1\) and, for any point \(x \in X\) and closed set \(B\) such that \(x \notin B\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(B \subseteq V\). A topological space \(X\) is normal iff it is \(T_1\) and, for any disjoint closed sets \(A\) and \(B\), there exist disjoint sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\). Lemma 31.1 Let \(X\) be a \(T_1\) space. \(X\) is regular if and only if, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\). \(X\) is normal if and only if, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\). Proof If \(X\) is regular then, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such th...
Read more