Definition of a topological space via neighborhoods

Question

By definition we have:

A subset $V$ of a topological space $X$ is a neighborhood of a set $S$ contained in $X$ if and only if $V$ is a neighborhood of every points in $S$.

On the other hand from the definition of a topological space (from the axiomatization due to Felix Hausdorff) using the concept of neighborhoods we have that:

If $V$ is a neighborhood of a set $S$ then it contains a neighborhood $U$ of $S$ such that $V$ is a neighborhood for each point contained in $U$

I am asking if the first claim does have any proof or is true just by the definition from the second one?

Answer 1

In your question, your second box is not a "definition". In order to "prove" the statement in the first box, one needs a definition:

Definition of neighborhood: Given a topological space $X$, a subset $S \subset X$, and a subset $V \subset X$, to say that $V$ is a neighborhood of $S$ means that there exists an open subset $U \subset X$ such that $S \subset U \subset V$. Given a point $x \in X$ and a subset $V \subset X$, to say that $V$ is a neighborhood of $x$ means that $V$ is a neighborhood of $\{x\}$; equivalently, there exists an open subset $U \subset X$ such that $x \in U \subset V$.

Here's a proof of the first statement.

For the "only if" direction, suppose that $V \subset X$ is a neighborhood of $S$, so there exists an open subset $U \subset X$ such that $S \subset U \subset V$. Choose $x \in S$. It follows that $x \in U \subset V$. So $V$ is a neighborhood of $x$.

For the "if" direction, suppose that $V$ is a neighborhood of each $x \in S$, so there exists an open $U_x \in X$ such that $x \in U_x \subset V$. It follows that the set $U$ which is defined by $U = \bigcup_{x \in S} U_x$ is an open subset of $X$, and clearly $S \subset U \subset V$, and therefore $V$ is a neighborhood of $S$.