The Induced Topology on a Subset of a Topological Space

# The Induced Topology on a Subset of a Topological Space

Definition: Let $(E, \tau)$ be a topological space and let $A \subseteq E$. The Induced Topology on $A$ (or the Subspace Topology on $A$) is the topology on $A$ given by $\{ A \cap U : U \in \tau \}$. |

It is easy to see that the induced topology on $A$ is indeed a topology on $A$.

Proposition 1: Let $(E, \tau)$ be a topological space and let $A \subseteq E$. Then:(1) The closed sets in $A$ are sets of the form $A \cap C$ where $C$ is a closed set in $E$.(2) If $x \in A$, then every neighbourhood of $x$ in $A$ is of the form $A \cap V$ where $V$ is a neighbourhood of $x$ in $E$. |

Proposition 2: Let $(E, \tau)$ be a topological space and let $A \subseteq E$. Then:(1) If $E$ is a Hausdorff space then $A$ is a Hausdorff space.(2) If $E$ is first countable, then $A$ is first countable.(3) If $E$ is metrizable, then $A$ is metrizable. |

**Proof of (1):**Let $x, y \in A$ with $x \neq y$. Since $A \subseteq E$, there exists neighbourhoods $U, V \subseteq E$ such that $x \in U$, $y \in V$, and $U \cap V$. But then $A \cap U$ and $A \cap V$ are neighbourhoods on $x$ contained in $A$ such that $x \in A \cap U$, $y \in A \cap V$, and:

\begin{align} \quad (A \cap U) \cap (A \cap V) = \emptyset \end{align}

- So $A$ is Hausdorff. $\blacksquare$

**Proof of (2):**Let $x \in A$. Since $A \subseteq E$, there exists a countable base of neighbourhoods of $x$ contained in $E$, denote it by $\{ U_n \subseteq E : n \in \mathbb{N} \}$. Then $\{ A \cap U_n \subseteq A : n \in \mathbb{N} \}$ is a countable base of neighbourhoods of $x$ contained in $A$. So $A$ is first countable. $\blacksquare$

**Proof of (3):**If $E$ is metrizable and if $d : E \times E \to [0, \infty)$ is a metric on $E$ for which the open sets of $E$ are unions of open balls $\{ V(x, \epsilon) : x \in E, \epsilon > 0 \}$, then it can be shown that the restriction metric $d_A : A \times A \to [0, \infty)$ generates the induced topology on $A$. $\blacksquare$