Induced topology on subset are equal got a huge dubt on a fact that was presented today to me in class. I would really appreciate if you could help me out with this!
Today in class the teacher mentioned the following:
Lets consider three subsets $W \subset Y \subset X$. Let $\tau$ be a topology on X and let $\hat{\tau}$  be a topology induced by $\tau$ in Y. Then the topologies induced by $\tau$ and $\hat{\tau}$ on $W$ are equal.
As I said, we are just starting the course on some kind of intruductory topology and I've searched for this fact online but couldnt find anything. Any help would be great.
Thanks in advance!
 A: $U  \subseteq W$ is open in the $X$ induced topology if and only if there is an open set $V \subseteq X$ such that
$$
U = V \cap W
$$
Note that
$$
V \cap W = (V \cap Y) \cap W
$$
since $W \subseteq Y$, and so if $U$ is open in the $X$ induced topology, it is open in the $Y$ induced topology and vice versa, since every open set in $Y$ can be written in the form $V \cap Y$ for some $V$ open in $X$.
A: This is a special case of the "transitive law of initial topologies", that I prove and describe here.
But this special case is also easy to see without abstract machinery if you prefer:
Let $U \subseteq W$ be open in the subspace topology on $W$ wrt $X$. Then $U = O \cap W$ for some $O$ open in $X$.
Then $O \cap Y$ is open in $Y$ by definition so $O \cap Y \cap W$ is open in $W$ when we give $W$ the subspace topology wrt $Y$. But $Y \cap W= W$ so $O \cap Y \cap W = O \cap W=U$. SO $U$ is also open wrt $Y$.
Now let $U \subseteq W$ be open in the subspace topology on $W$ wrt $Y$. Then $U= V \cap W$ where $V$ is open wrt $Y$. But as $Y$ has the subspace topology wrt $X$ we know that $V = O \cap Y$ where $O$ is open in $X$. But then $U = V \cap W = (O \cap Y) \cap W = O \cap (Y \cap W) = O \cap W$ and $U$ is also open in the subspace topology on $W$ wrt $X$.
So "subspace topology on $W$ wrt $X$" is the same topology as "subspace topology on $W$ wrt $Y$".
