Relative topology theorem hunt.

Question

Let $(X,T)$ be a topological space and let $Y\subset X$

1) Assume that Y is an open subset of X. Find and prove a theorem that describes the topology $T_Y$ in simpler terms.

Originally i thought the statement that i was looking for was $B_Y= \{ B\cap Y | B \in B' \}$ where $B'$ is a basis for T and $B_Y$ was a basis for the relative topology. i was able to prove this.

the problem is the next question is the following:

Assume that Y is a closed subset of X. Find and prove an analogous theorem to the previous one.

so my statement isnt the right one cause my proof didnt seem to care if Y was open or closed?

Answer 1

I suspect the answers they have in mind are the following:

If $Y$ is open, then $T_Y = \{U\mid U\in T, U\subseteq Y\}$.

If $Y$ is closed, then $T_Y = \{Y\backslash U \mid U\in T, U\subseteq Y\} = \{C\mid C \text{ is closed relative to }T, C\subseteq Y\}$.

Answer 2

Relative topologies don't care if the subset is open or closed. The definition is exactly the same in any case. So your answer to part (1) is correct, and if you copied that answer word-for-word to part (2) it would still be correct.

Answer 3

For me what you gave is not a theorem, it is the definition of the induced topology. I guess that is a matter of taste. Then a theorem would be

Theorem. The induced topology on $Y\subset X$ is the coarsest topology such that the inclusion $Y\hookrightarrow X$ is a continuous map.

which is valid whether $Y$ is open, closed, or none of these.

Answer 4

I think probably, what the question is looking for is along this lines: note that if $Y$ is open, then for any open $U \subseteq X$, then $U \cap Y$ is open in $X$. Therefore, $T_Y \subseteq T_X \cap P(Y) = \{ S \subseteq Y \mid S \in T_X \}$. In fact, it turns out that $T_Y = T_X \cap P(Y)$ - I'll leave proving the reverse inclusion to you.

Then, for (2), you could come up with a similar statement using the fact that if $Y$ is closed, then for any $F \subseteq X$ which is closed, we have $F \cap Y$ is also closed in $X$.