The usual topology for $\mathbb{R}$

Question

I am working through some exercises in the dover book "Elementary Point Set Topology"

I am given the following axioms for a "system of neighborhoods" at x.

Let $X$ be a set and for each $x\in X$ define $\mathcal{U}_x=\{U(x)\}$ as a family of subsets of $X$ associated with $x$ such that

1. $x\in U(X)$ for each $U(x)\in \mathcal{U}_x$
2. If $U(x)\subset V$ for some $U(x)$, then $V\in \mathcal{U}_x$.
3. If $U,V\in\mathcal{U}_x$, then $U\cap V\in \mathcal{U}_x$.
4. If $U\in\mathcal{U}_x$, then there exists $V\in\mathcal{U}_x$ such that if $y\in V$, then $U\in\mathcal{U}_y$.

I am having a hard time interpreting the meaning of the fourth axiom. Please critique the following proof from an excercise in the book.

Let $\mathbb{R}$ be the real line and define $(a,b)=\{x|a<x<b\}$ and define $\mathcal{U}_x=\{U|x\in(a,b) \subseteq U \ \text{for some}\ a,b\in\mathbb{R},a<b\}$,

then $\mathcal{U}_x$ is a neighboorhood system at $x$.

Proof.

1. $x\in U(x)$ for each $U(x)\in\mathcal{U}_x$ by definition.
2. Suppose that $U(x)\subset V$ for some $U(x)\in\mathcal{U}_x$. Then $x\in (a,b)\subset U\subset V$ for some $a,b\in\mathbb{R}$. Hence, $V\in \mathcal{U}_x$.
3. Suppose that $U,V\in\mathcal{U}_x$. Then $x\in(a_1,b_1)\subset U$ and $x\in(a_2,b_2)\subset V$ for some $a_1,b_1,a_2,b_2\in\mathbb{R}$. Let $(a,b)=(min\{a_1,a_2\},min\{b_1,b_2\})$. Then $x\in(a,b)\subset U\cap V$. (Here I am not sure if Im correctly interpereting the defininition of $\mathcal{U}_x$ for the usual topology of $\mathbb{R}$.)
4. Suppose that $U(x)\in\mathcal{U}_x$. Choose $V=(a,b)$. Then $V\in \mathcal{U}_x$ since $x\in(a,b)\subset V$. Let $y\in V$. Then $y\in(a,b)\subset V$. But $(a,b)\subset U$ hence $y\in(a,b)\subset U$. Therefore, $U\in \mathcal{U}_y$.

Answer 1

This question and its answers might help you understand better why this 4th axiom is needed. My answer here proves the equivalence between neighbourhood spaces and topological spaces, and also shows the use of the 4th axiom.

As to your proof for the case of the real line: it's fine as to axiom 4. In terms of "neighbourhood spaces" it shows that the sets $(a,b)$ containing $x$ are the open neighbourhoods of $x$ (while sets containing open intervals containing $x$ are all neighbourhoods of $x$).

In axiom 3 (intersection), use $a_3 = \max(a_1, a_2), b_3 = \min(b_1,b_2)$ for the intersection (draw a picture): $a_1 < x$ and $a_2 < x$ iff $\max(a_1, a_2) < x$ and $x < b_1$ and $x < b_2$ iff $x < \min(b_1,b_2)$ are both easily verified.