Subspace (induced) topology

Definition: Subspace (induced) topology

Given a topological space $(X,\mathcal{T})$ and a subset $A\subseteq X$, the subspace or induced topology on $A$ is defined by the collection

$${\mathcal{T}}_A=\{A\cap U|U\in \mathcal{T}\}.$$

In other words, open sets in $A$ (i.e., a subset of $A$ that is open in the subset topology) are obtained by intersecting $A$ with open sets of $X$ (i.e., an open set in $(X,\mathcal{T}$)).

Theorem: Universal property of the subspace topology

Let $f:X\to Y$ be a function between topological spaces, and $S\subset Y$ be a subset equipped with the subspace topology. If $f(X)\subset S$ and $g:X\to S$ is the function obtained by restricting the codomain of $f$, then $g$ is continuous if an only if $f$ is continuous.

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See also: Product topologies and subspaces

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Basis for a subspace topology §

Lemma: Basis for a subspace topology

If $\mathcal{B}$ is a basis for the topology $\mathcal{T}$ of $X$ and $A\subseteq X$ is any subset, then

$${\mathcal{B}}^{\prime }=\{A\cap U|U\in \mathcal{B}\}$$

is a basis for the subspace topology on $A$.

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Closed sets in subspace topologies §

Theorem (Munkres 17.2): Closed sets in a subspace topology

Let $Y\subset X$ be a subspace. Then a set $A\subset Y$ is closed in $Y$ if and only if it is the intersection of $Y$ with a closed set in $X$.

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Review §

• Give an example where a subset is open in $A\subset X$ but not in $X$.
• Prove the lemma about a basis for a subspace topology. ⭐
• (Topology HW1, P1) Show that if $X$ is a topological space and $A\subset X$ has the property that for all $x\in A$, there exists a open set $U$ containing $x$ such that $U\subset A$, then $A$ is open in $X$.
• (Topology HW1, P2) Show that if $Y$ is a subspace of $X$ and $A$ is a subset of $Y$, then the subset topology $A$ inherits as a subspace of $Y$ is the same as the subspace topology $A$ inherits as a subspace of $X$.