subspace of a subspace
Theorem 1.
Suppose are sets and is a
topological space with topology .
Let be the subspace topology in given by ,
and let be the subspace topology in given by
, and let be the subspace topology in
given by . Then .
Proof.
Let , then there is by the definition of the subspace topology an open set such that . Now and therefore . But since , we have , so and thus .
To show the reverse inclusion, take an open set . Then there is an open set such that . Furthermore, there is an open set such that . Since , we have
so and thus .
Together, both inclusions yield the equality . ∎
Title | subspace![]() |
---|---|
Canonical name | SubspaceOfASubspace |
Date of creation | 2013-03-22 15:17:53 |
Last modified on | 2013-03-22 15:17:53 |
Owner | matte (1858) |
Last modified by | matte (1858) |
Numerical id | 5 |
Author | matte (1858) |
Entry type | Theorem |
Classification | msc 54B05 |