subspace of a subspace
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 of a subspace|
|Date of creation||2013-03-22 15:17:53|
|Last modified on||2013-03-22 15:17:53|
|Last modified by||matte (1858)|