Subspace Topology is Initial Topology with respect to Inclusion Mapping

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

Let $Y$ be a non-empty subset of $X$.

Let $\iota: Y \to X$ be the inclusion mapping.

Let $\tau_Y$ be the initial topology on $Y$ with respect to $\iota$.

Then $\left({Y, \tau_Y}\right)$ is a topological subspace of $\left({X, \tau}\right)$.

That is:
 * $\tau_Y = \left\{{U \cap Y: U \in \tau}\right\}$

Proof
By Initial Topology with respect to Mapping equals Set of Preimages, it follows that:
 * $\tau_Y = \left\{{\iota^{-1} \left({U}\right): U \in \tau}\right\}$

From Preimage of Subset under Inclusion Mapping, we have:
 * $\forall S \subseteq X: \iota^{-1} \left({S}\right) = S \cap Y$

Hence the result.