Mathematics Prelims

January 24, 2009

The Discrete, Trivial, and Subspace Topologies

Filed under: Topology — cjohnson @ 5:51 pm

Given a set X, there are always at least two topologies that can be placed on X: the discrete topology and the trivial topology.  The discrete topology is the topology where every set is open; the collection of open sets in the topology is simply the collection of all sets, 2^{X}.  If f : X \to Y is a map between topological spaces where X has the discrete topology, then f will be continuous, regardless of how it maps the elements of X to Y.  The trivial topology is the bare minimum for a topology: the empty set and the entire set X.  For any map f : Z \to Y, then, f will be continuous as f^{-1}(\emptyset) = \emptyset and f^{-1}(Y) = Z.  Regardless of the topology on Z, those two sets will be open, so f is continuous.

Suppose now that (X,\tau) is a topological space where \tau is any topology.  If Y \subseteq X we can place a topology on Y by using the topology of X.  Let \tau_2 be a collection of subsets of Y where Z \in \tau_2 (Z is an open subset of Y) if Z = Y \cap \mathcal{O} for some \mathcal{O} \in \tau.  We call \tau_2 the subspace topology of Y.  For instance if we take \mathbb{R} with the standard topology, we can place a topology on {}[0, 1] using this subspace topology.  In such a case, sets like {}[0, 1/2) are considered open in this topology since {}[0, 1/2) = [0, 1] \cap (-1/2, 1/2), even though this is not considered an open subset of \mathbb{R}.

If A and B are topological spaces and f : A \to B is continuous, we’d like to know when f is continuous when restricted to a subspace C of A.  We can show that if define D as the image of C under f, and assign D the subspace topology in B, then f|_C : C \to D is continuous.  Let U be an open subset of D.  Then U = D \cap \mathcal{O} for some open subset \mathcal{O} of B.  We know f|_C^{-1}(U) = f^{-1}(D \cap \mathcal{O}) = f^{-1}(D) \cap f^{-1}(\mathcal{O}).  By definition f^{-1}(D) = C and f^{-1}(\mathcal{O}) is an open subset of A by the continuity of f.  That is, f|_C^{-1}(U) = C \cap f^{-1}(\mathcal{O}) is an open subset of C with the subspace topology, so the restriction of f to C is continuous.

In the case of \mathbb{Z}, the subspace topology inherited from \mathbb{R} is the same as the discrete topology on \mathbb{Z}.  Let z \in \mathbb{Z}.  Then \{z\} is an open set with the subspace topology as \{z\} = \mathbb{Z} \cap (z - 1/2, z + 1/2).  This gives that every singleton set is an open set and since any set will be a union of singletons, any set will be open.


1 Comment »

  1. […] The Trivial, Discrete, and Subspace Topologies […]

    Pingback by Topology, Geometry & Dynamics | Topology, Geometry & Dynamics — September 24, 2010 @ 4:56 pm | Reply

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