subspace of a subspace
Theorem 1.
Suppose are sets and is atopological 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 subspacetopology 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 , wehave
so and thus .
Together, both inclusions yield the equality .∎