Subspace Topology is Initial Topology with respect to Inclusion Mapping

Theorem

Let $\struct {X, \tau}$ 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 $\struct {Y, \tau_Y}$ is a topological subspace of $\struct {X, \tau}$.

That is:

$\tau_Y = \set {U \cap Y: U \in \tau}$

Proof

By Initial Topology with respect to Mapping equals Set of Preimages, it follows that:

$\tau_Y = \set {\iota^{-1} \sqbrk U: U \in \tau}$

From Preimage of Subset under Inclusion Mapping, we have:

$\forall S \subseteq X: \iota^{-1} \sqbrk S = S \cap Y$

Hence the result.

$\blacksquare$