Munkres Topology Proof Check

Question

I'm struggling a bit intuitively understanding this topology introduction by Munkres (second edition). I'm on chapter 2 (self study, hobbyist) and working through problem 16.1. I think I have a solution, but I'm not confident in it. Heres the question:

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.

My solution: Take a basis element of the basis of the topology $T_{A_{y}}$, the subspace of A under Y, say $V = A \cap U$ with $U$ open in $Y$. $V$ is an element of the topology of $Y$, which is a subspace of X, so this element can be expressed as $V = Y\cap U_x$ with $U_x$ an element of the subspace topology of Y under X. This means we can express $V$ as $V = A\cap (Y \cap U_x )$. Since $(Y \cap U_x)$ is open in the topology of $X$, $V$ is an element of the topology of the subspace of $A$ as it inherits from $X$.

Here are my questions:

1. is this correct?
2. I'm implicitly assuming that if I take a basis element of $T_{A_{Y}}$, if I express it in the basis of a topology under $Y$, then it must be expressable (if thats a word) as a basis element of the subspace of $Y$ under $X$. Must this be the case? Could it not be but be expressable under some other topology?

Answer 1

It's worth being explicit about the topologies you're working with here. We are assuming $X$ is a topological space, so there is a topology $T$ on $X$ with open sets $U$.

If $Y$ is a subspace of $X$, then the subspace topology on $Y$ is given by $$T'=\{Y\cap U: U\in T\}.$$

Next, there is a topology defined on $Y$, with $A$ being a subset of $Y$, so we can define the subspace topology on $A$ that it inherits as a subspace of $Y$ as $$T''=\{A\cap V: V\in T'\}.$$ Finally, since $X$ is a topological space with $A\subseteq X$, we can define the subspace topology on $A$ that it inherits as a subspace of $X$ as $$T'''=\{A\cap U: U\in T\}.$$

Your goal is to show $T''=T'''$. Try attacking this with an element chasing argument by showing $T''\subseteq T'''$ and $T'''\subseteq T''.$ Notice no bases are mentioned. It's usually ill-advised to assume the existence of a basis for a topology unless you're referring to the topology itself, which is always a basis. But in that case, you're just doing the element chasing argument I suggested since the basis is the topology itself, with elements the open sets.

So, just show open sets of the correct form are contained in each topology.

A hint to get you started:

You started to show $T''\subseteq T'''$. If $A\cap V$ is an element of $T''$, then since $V=Y\cap U$ for some $U\in T$, we may write $A\cap V = A\cap (Y\cap U)=(A\cap Y)\cap U=A\cap U$, since $A$ is a subset of $Y$.