Topological Subspace is Topological Space/Proof 2

Theorem

Let $\struct {X, \tau}$ be a topological space.

Let $H \subseteq X$ be a non-empty subset of $X$.

Let $\tau_H = \set {U \cap H: U \in \tau}$ be the subspace topology on $H$.

Then the topological subspace $\struct {H, \tau_H}$ is a topological space.

Proof

Follows directly from Subspace Topology is Initial Topology with respect to Inclusion Mapping.

$\blacksquare$

Also see

• Definition:Topological Subspace