Question

Topology question.

Answer 1

Suppose \(\mathcal{T}_1\) and \(\mathcal{T}_2\) are topologies on \(X\) s.t. \(\mathcal{T}_1\) is strictly finer than \(\mathcal{T}_2\). Suppose \(Y\subset X\). Why is it NOT always the case that the subspace topology on \((Y,\mathcal{T}_1)\) is strictly finer than \((Y,\mathcal{T}_2)\)?

I understand that it is finer but I don't understand why it may not be strict.

@eliassaab

Answer 2

Take \[Y= \emptyset \subset X
\]

Answer 3

oh

Answer 4

right

Answer 5

Wait would we not get the trivial topology as a subspace?

I thought we were trying to show that we could have Y proper in X where topology on X is the same as the topology on Y?

Answer 6

You can also take Y to be finite subset of X

Answer 7

So take two topologies, and the null set of X, we get the same subspace topology under both topologies, and this is just the trivial topology in both situations, even though one topology is strictly finer than the other.

Got it!