J. Munkres. Topology (2013) Chapter 2: Topological Spaces and Continuous Functions. 16: The Subspace Topology

Definition

Let \(X\) be a topological space. Let \(Y\) be a subset of \(X\). The subspace topology on \(Y\) is \(\{ U \cap Y \mid U \text{ open in } X \}\). We call \(Y\) with this topology a subspace of \(X\).

Lemma 16.1

Let \(X\) be a topological space. Let \(Y\) be a subspace of \(X\). Let \(\mathcal{B}\) be a basis for the topology on \(X\). Then \(\{B \cap Y \mid B \in \mathcal{B}\}\) is a basis for the subspace topology on \(Y\).

Lemma 16.2

Let \(X\) be a topological space. Let \(Y\) be a subspace of \(X\). Let \(U \subseteq Y\). If \(U\) is open in \(Y\) and \(Y\) is open in \(X\) then \(U\) is open in \(X\).

Theorem 16.3

Let \(X\) and \(Y\) be topological spaces. Let \(A\) be a subspace of \(X\) and \(B\) a subspace of \(Y\). Then the product topology on \(A \times B\) is the same as the topology \(A \times B\) inherits as a subspace of \(X \times Y\).

Example 2

It is not true in general that, if \(X\) is a linearly ordered set with the order topology and \(Y \subseteq X\), then the order topology on \(Y\) agrees with the subspace topology on \(Y\). Take \(X = \mathbb{R}\) and \(Y = [0,1) \cup \{2\}\). Then the set \(\{2\}\) is open in the subspace topology but not in the order topology on \(Y\).

Example 3

The ordered square is the set \([0,1]^2\) under the dictionary order.

Definition

Let \(X\) be a linearly ordered set and \(Y \subseteq X\). Then \(Y\) is convex iff, for all \(a,b \in Y\) and \(c \in X\), if \(a < c < b\) then \(c \in Y\).

Theorem 16.4

Let \(X\) be a linearly ordered set with the order topology. Let \(Y\) be a convex subset of \(X\). Then the order topology on \(Y\) is the same as the subspace topology.

Exercise 1

If \(Y\) is a subspace of \(X\) and \(Z\) is a subspace of \(Y\), then the subspace topology on \(Z\) as a subspace of \(X\) is the same as the subspace topology in \(Z\) as a subspace of \(Y\).

Definition

Let \(X\) and \(Y\) be topological spaces and \(f : X \rightarrow Y\). Then \(f\) is an open map iff, for every open set \(U\) in \(X\), the set \(f(U)\) is open in \(Y\).

(Strange that Munkres introduces this definition before continuous functions, and in an exercise!)

Exercise 4

Given topological spaces \(X\) and \(Y\), the projections \(\pi_1 : X \times Y \rightarrow X\) and \(\pi_2 : X \times Y \rightarrow Y\) are open maps.

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