1. Topological Spaces (20/8/2025).
  2. Bases (27/8/2025).
  3. Metrics (3/9/2025).
  4. Interval Topology (17/9/2025).
  5. Disjoint Unions (12/11/2025).

Interval Topology

due by Wednesday, Sep 17, 2025

Let $(A, \leq)$ be a totally ordered set. For $p, q\in A$, we define the open intervals $(p, q)$, $(-\infty, p)$ and $(p, \infty)$ as $(p, q) = \{x\in A : p < x, x < q\}$, $(-\infty, p) = \{x\in A : x < p\}$, and $(p, \infty) = \{x\in A : p < x\},$ where in the first case we assume $p < q$.

  1. Show that if $A$ has at least two elements, the collection of subsets of $A$ given by $\{(p, q): p,q \in A, p <q \}\cup \{(-\infty, p): p\in A\}\cup \{(p, \infty): p \in A\}$ forms a base for a topology on $A$. We call this the interval topology.
  2. Suppose $A$ and $B$ are totally ordered sets with at least two elements and $f: A \to B$ is a strictly increasing surjective function. Show the $f$ is continuous with respect to the interval topologies on $A$ and $B$.
  3. Let $A$ be a totally ordered set and $B \subset A$ be a subset with at least two elements. We consider the total order on $B$ obtained by restricting the order on $A$. Prove or disprove: The interval topology on $B$ is the subspace topology with respect to the interval topology on $A$.