Are these sets representing the subspace topology equal? 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? $$\mathcal T'= \{U\in \mathcal T \,|\,U\subseteq \Omega'\}$$

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.)
