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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!

user1trill
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2 Answers2

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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 $Y$".

Henno Brandsma
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$U \subseteq W$ 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 $X$.

Holmes
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