Are these sets representing the subspace topology equal?

Question

I have a question regarding the representation of the subspace topology.
Let $(\Omega, \mathcal T)$ be a topological space, $\Omega' \subseteq \Omega$ and $\mathcal T' \subseteq \mathcal T$ the subspace topology with respect to $\mathcal T$. Then, by definiton $$\mathcal T' = \{U\cap\Omega' \,|\, U\in \mathcal T\}.$$ My question: Is the following representation equal to the first one? $$\mathcal T'= \{U\in \mathcal T \,|\,U\subseteq \Omega'\}$$

Answer 1

The second definition is not correct in general.

The subspace topology $\Omega '$ is composed by the intersection of the open sets of $\Omega$ with $\Omega'$, hence is possible that there are $U\in\mathcal T$ that are not subsets of $\Omega'$ and that at the same time we need them to define the subtopology (the open sets) of $\Omega'$.

By example take the topological subspace of $\Bbb R$ defined by $A:=[0,1)\cup(2,3]$. Then from the open sets $(-1,1)$ and $(2,4)$ of (the standard topology of) $\Bbb R$ we have that

$$A\cap (-1,1)=[0,1),\quad A\cap (2,4)=(2,3]\tag{1}$$

Hence $[0,1)$ and $(2,3]$ are open sets in the topological subspace $A$ (by your first definition of topological subspace).

But if we follow your second "definition" of topological subspace we cannot define the open sets in $(1)$ because is clear that $[0,1)$ and $(2,3]$ are not open sets in (the standard topology of) $\Bbb R$.

Answer 2

No, for a counterexample, consider the topological space $\Omega = \mathbb{R}$ with the usual order or metric topology, and $\Omega' = [0,1]$. Then $(1/2, 1]$ is in the subspace topology because it is equal to $(1/2, 3/2) \cap \Omega'$; however, it is not in $\mathcal{T}$.

In fact, in general the only way the two sets can be equal is if $\Omega'$ is itself open in $\Omega$, since $\Omega \in \mathcal{T}$ implies $\Omega' = \Omega \cap \Omega' \in \mathcal{T}'$. (And in this case it's easy to show your equality.)