Topological Subspaces
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Topological Subspaces

Recall from the Initial Topologies page that if $X$ is a set, $\{ Y_i : i \in I \}$ is a collection of topological spaces, and $\{ f_i : X \to Y_i : i \in I \}$ is a collection of maps, then the initial topology induced by $\{ f_i : i \in I \}$ on $X$ is the topology $\tau$ which makes $f_i : X \to Y_i$ continuous for all $i \in I$.

We will now look at a very important type of topology known as a subspace topology.

Definition: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. The Topological Subspace or simply Subspace topology on $A$ is the topology $\tau_A = \{ A \cap U : U \in \tau \}$.

From the definition above it is not clear whether or not $\tau_A$ indeed forms a topological space $(A, \tau_A)$. The following theorem will show that the subspace topology $\tau_A$ is in fact a topology on $A$.

Theorem 1: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then the collection $\tau_A = \{ A \cap U : U \in \tau \}$ is a topology on $A$.
  • Proof: Let $A$ be a subset of the topological space $(X, \tau)$ and consider the inclusion function $i : A \to X$ defined for all $a \in A$ by:
\begin{align} \quad i(a) = a \end{align}
  • That is, for all $a \in A$ we have that $i(a) = a \in X$. The initial topology on $A$ induced by $i$ is the coarsest topology on $A$ that makes $i : A \to X$ continuous. This topology has subbasis:
\begin{align} \quad \{ i^{-1}(U) : U \in \tau \} \end{align}
  • Since $U$ is an open set in $X$ and $A \subseteq X$, we have that $i^{-1}(U) = A \cap U$ for all $U \in \tau$.
  • So $\tau_A = \{ A \cap U : U \in \tau$, i.e., the subspace topology $\tau_A$ is indeed a topology since it equals the initial topology on $A$ induced by the inclusion map $i$. $\blacksquare$
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