subspace topology in a metric space

Theorem 1.

Suppose X is a topological spaceMathworldPlanetmath whose topology is induced by a metric d, and suppose YX is a subset. Then the subspace topology in Y is the same as the metric topologyMathworldPlanetmath when by d restricted to Y.

Let d:Y:Y be the restriction of d to Y, and let

Br(x) = {zX:d(z,x)<r},
Br(x) = {zY:d(z,x)<r}.

The proof rests on the identity


Suppose AY is open in the subspace topology of Y, then A=YV for some open VX. Since V is open in X,


for some ri>0, xiX, and

A = {YBri(xi):i=1,2,}
= {Bri(xi):i=1,2,}.

Thus A is open also in the metric topology of d. The converse direction is proven similarly.

Title subspace topology in a metric space
Canonical name SubspaceTopologyInAMetricSpace
Date of creation 2013-03-22 15:17:44
Last modified on 2013-03-22 15:17:44
Owner matte (1858)
Last modified by matte (1858)
Numerical id 5
Author matte (1858)
Entry type Theorem
Classification msc 54B05