closed set in a subspace


In the following, let X be a topological spaceMathworldPlanetmath.

Theorem 1.

Suppose YX is equipped with the subspace topology, and AY. Then A is closed (http://planetmath.org/ClosedSet) in Y if and only if A=YJ for some closed setPlanetmathPlanetmath JX.

Proof.

If A is closed in Y, then YA is open (http://planetmath.org/OpenSet) in Y, and by the definition of the subspace topology, YA=YU for some open UX. Using properties of the set differenceMathworldPlanetmath (http://planetmath.org/SetDifference), we obtain

A = Y(YA)
= Y(YU)
= YU
= YU.

On the other hand, if A=YJ for some closed JX, then YA=Y(YJ)=YJ, and so YA is open in Y, and therefore A is closed in Y. ∎

Theorem 2.

Suppose X is a topological space, CX is a closed set equipped with the subspace topology, and AC is closed in C. Then A is closed in X.

Proof.

This follows from the previous theorem: since A is closed in C, we have A=CJ for some closed set JX, and A is closed in X. ∎

Title closed set in a subspace
Canonical name ClosedSetInASubspace
Date of creation 2013-03-22 15:33:32
Last modified on 2013-03-22 15:33:32
Owner yark (2760)
Last modified by yark (2760)
Numerical id 9
Author yark (2760)
Entry type Theorem
Classification msc 54B05