Understanding a lemma regarding subspace topology [duplicate]

Question

I am learning topology and am struggling with some of the concepts of open sets regarding subspaces.

The problem I am working on says, that Y is a subspace of X, and A is a subset of Y, then show that the topology A inherits as a subspace of Y is the same topology it inherits as a subspace from X.

now what I am trying to prove is that A is a basis for the subspace topology of Y of X, and A is the basis for the subspace topology on X, then these two topologies are the same.

Here is the lemma I am struggling to utilize , it says, "Let Y be a subspace of X, if U is open in Y and Y is open in X, then U is open in X."

Since A is a subset of Y, does this imply that A is open in Y? (i think so). Y is open in X, so does this imply that A is open in X? (I think so)

If A is open in Y, and A is open in X, does this mean that A is open in $Y \bigcap U$ for $U \subset X$ ? (This I am not sure of).

Further, assuming my intuition is right, if A is open in $Y \bigcap U$ does this mean for $x \in Y \bigcap U$ that $x \in A \bigcap U$ since A is open in both Y and X?

this yields my needed argument saying that $A \bigcap U \subset Y \bigcap U$ this can then be shown that $\mathbb{B}_{A}$ is a basis for both topologies. my question is that just because A is a subset of Y and A is open in X, and $x \in Y \bigcap U$ , I am not sure if this implies that $x \in A \bigcap U$.

Thank you again so very much and I will greatly appreciate any clarification.

Answer 1

Answer on the problem mentioned in second alinea.

Let it be that $X$ is a set equipped with a topology $\tau$, and let $A\subseteq Y\subseteq X$.

Then $Y$ inherits a subtopology $\tau_{Y}=\left\{ Y\cap U\mid U\in\tau\right\} $ and likewise $A$ inherits a subtopology $\tau_{A}=\left\{ A\cap U\mid U\in\tau\right\} $.

Looking at $A$ as a subspace of $Y$ it also inherits a subtopology $\tau'_{A}=\left\{ A\cap V\mid V\in\tau_{Y}\right\} $ and to be shown is that $\tau_{A}$ and $\tau'_{A}$ coincide.

If $O\in\tau'_{A}$ then $O=A\cap V$ for some $V\in\tau_{Y}$. From $V\in\tau_{Y}$ it follows that $V=Y\cap U$ for some $U\in\tau$. Then $O=A\cap Y\cap U=A\cap U$ showing that $O\in\tau{}_{A}$.

Proved is now that $\tau'_{A}\subseteq\tau_{A}$.

Conversely let $O\in\tau_{A}$. Then $O=A\cap U$ for some $U\in\tau$. Now note that we can also write $O=A\cap Y\cap U=A\cap W$ where $W:=Y\cap U\in\tau_{Y}$. This shows that $O\in\tau'_{A}$.

Proved is now that $\tau{}_{A}\subseteq\tau'_{A}$ and you are ready.

Answer 2

A basis is a collection of sets, so for a mere subset, you have to show that the subset is a collection of sets, not just a single set before showing that a subset can be a basis.

The second part of my confusion, assuming that A is open in Y and X, is that if one were to show that A were open in $Y \bigcap U$ where U $\subset X $ , one needs to show that A is contained in U. both of these I do not have and can not assume.

thank you very much.