Questions regarding subspace topology theorems for either open sets vs neighborhood

Question

For the following theorem:

Theorem 1: Let $(A,T_A)$ be a subspace of a topological space $(X,T)$, let $U\subset A$, and let $a\in A$. then $U$ is a neighborhood of $a$ with respect to $T_A$ if and only if there is a neighborhood $V$ of $a$ with respect to $T$ such that $U=A\cap V.$

If the phrase "$U$ is an open set such that $a\in U$" means "$U$ is a neighborhood of $x$", then don't I prove the above theorem the same way as I prove about open sets in subspace topology stated as theorem 2:

Theorem 2: Let $(A,T_A)$ be a subspace of a topological space $(X,T)$, let $U\subset A$, then $U$ is an open set of $a$ in the subspace topology $T_A$ if and only if there is an open set $V$ of $X$ containing $a$ such that $U=A\cap V.$

The reason I ask is because proving facts about neighborhood of a point $x$ in the context of subspace topology is not exactly the same or straightforward as proving about open sets. I am just wondering if we know about theorem 2, is theorem 1 really necessary in the theory about subspace topology. When I mean necessary, are there any further results which would be easier having theorem 1 already proved.

Thank you in advance.

Answer 1

If we define neighbourhood as usual (see e.g. Wikipedia) and the subspace topology from $(X, \mathcal{T})$ on $A \subseteq X$ by

$$\{O \cap A: O \in \mathcal{T}\}$$

then "Theorem 2" is almost immediate: if $U \subseteq A$ is an open neighbourhood of $a \in A$, if $U$ is open in $A$ (so $U = V \cap A$ for some $V \in \mathcal{T}$, i.e. open in $X$) and $a \in U$ (so $a \in V$ too). So $V$ is an open neighbourhood of $a$, that, when intersected with $A$ gives $U$. The reverse is also obvious: if $V$ is an open neighbourhood (in $X$) of some $a \in A$, then $V \in \mathcal{T}$ so $U:=V \cap A$ is open in $A$ by definition of the subspace topology, and still $a \in U$ (definition of intersection).

Sometimes the set of open sets containing $a$ is denoted $\mathcal{O}(a)$ and we can write this succinctly (using subscripts for the space we are considering) as:

$$\forall a \in A: \mathcal{O}_A(a) = \{O \cap A: O \in \mathcal{O}_X(a)\}$$

and Theorem 1. is the extension of this to all neighbourhoods $\mathcal{N}(a)$ of points of $a$:

$$\forall a \in A: \mathcal{N}_A(a) = \{O \cap A: O \in \mathcal{N}_X(a)\}$$

and the proof is as easy as that of Thm. 2. : if $U$ is a neighborhood of $a \in A$ it contains by definition an open neighbourhood $O$ of $a$. By 2. this is of the form $O' \cap A$ where $O' \in \mathcal{O}_X(a)$. So we can just define $V= U \cup O'$ which is in $\mathcal{N}_X(a)$ as a superset of $O'$ and $V \cap A= (U \cap A) \cup (O' \cap A)=U \cup O = U$. Reversely, if $V$ is in $\mathcal{N}_X(a)$ and $a \in A$ we know that there is $O \in \mathcal{O}_X(a)$ with $O \subseteq V$ and then $a \in O \cap A \subseteq V \cap A$ so that $V \cap A \in \mathcal{N}_A(a)$ by definition of a neighbourhood of $a$.

Answer 2

If the phrase "$U$ is an open set such that $a \in U$" means "$U$ is a neighborhood of [$a$]"

It (usually) doesn't. The canonical definition of neighborhood in topology is (borrowing from your textbook's notation): subset $U \subseteq X$ is a neighborhood of point $a \in X$ if and only if $U$ contains an open set which contains $a$. Specifically, one doesn't require $U$ itself to be open.

Therefore Theorem 1 and Theorem 2, while very much akin to each other, are stating slightly different results. Actually Theorem 1 is a fairly trivial corollary of Theorem 2: can you see how?