# 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.