Exercise 1, Section 16 of Munkres Topology

Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

I rephrase this exercise to my taste.

If $(X,\mathcal{T}_{X})$, $Y\subseteq X$ with topology $\mathcal{T}_{Y}=\{Y \cap U|U\in \mathcal{T}_{X}\}$, $A\subseteq Y$, then $\mathcal{T}_{A}=\{A\cap R|R\in \mathcal{T}_{Y}\}$ ($A$ is a subspace of $Y$) is equal to $\mathcal{T}_{A}^\prime =\{A\cap S| S\in \mathcal{T}_{X}\}$ ($A$ is a subspace of $X$).

My attempt: let $M=(A\cap C) \in \mathcal{T}_{A}$. Then $C\in \mathcal{T}_{Y}$. That means $C=Y\cap D$, $D\in \mathcal{T}_{X}$. So $M=A\cap C= A\cap (Y\cap D)$. Since $A\subseteq Y$, we have $A\cap Y=A$. By associative law, $M=(A\cap Y)\cap D=A\cap D$, $D\in \mathcal{T}_{X}$. Thus $D\in \mathcal{T}_{A}^\prime$. Hence $\mathcal{T}_{A} \subseteq \mathcal{T}_{A}^\prime$.

Conversely, if $M\in \mathcal{T}_{A}^\prime$, then $M=A\cap C$, where $C\in \mathcal{T}_{X}$. Since $A\subseteq Y$, $Y\cap A=A$. So $M=(Y\cap A)\cap C=A\cap (Y\cap C)$, by associative and commutative laws. Since $C\in \mathcal{T}_{X}$, we have $Y\cap C\in \mathcal{T}_{Y}$. Thus, $M=A\cap (Y\cap C)\in \mathcal{T}_{A}$. Hence $\mathcal{T}_{A} \supseteq \mathcal{T}_{A}^\prime$. Is this proof correct?

Solution 1:

There is a problem with the first two sentences of your attempt. You wrote “let $M=A\cap C\in\mathcal T_A$. Then $C\in\mathcal T_Y$”. What is that $C$ in the first sentence and why does it belong to $\mathcal T_A$? It would be better if you had written “Let $M\in\mathcal T_A$. Then $M=A\cap C$, for some $C\in\mathcal T_Y$”.

The rest looks fine.