(related to Proposition: Properties of the Set of All Neighborhoods of a Point)

- By hypothesis, $(X,\mathcal O)$ is a topological space, $A\in X,$ and $\mathcal N(A)$ is the set of all neighborhoods of $A.$

- $X\in\mathcal N(A)$.
- Therefore, $\mathcal N(A)$ is not empty).

- $A\not\in\emptyset$.
- Therefore, $\emptyset$ is not an element of $\mathcal N(A).$

- Assume, $U,W\in\mathcal N(A).$
- By definition of neighborhoods, $A\in O_U\subset U$ and $A\in O_W\subset W$ with $O_U,O_W\in \mathcal O.$
- Since $X$ is a topological space, $\mathcal O$ contains the intersection $O_U\cap O_W.$
- On the other hand, $A\in O_U\cap O_W\subset U\cap W.$
- It follows that $U\cap W$ is a neighborhood of $A.$
- Thus, $U\cap W\in\mathcal N(A).$

- Assume, $U\in\mathcal N(A).$
- Thus, $A\in O\subset U$ for some $O\in \mathcal O.$
- But then $A\in O\subset U\subset W$ for any superset $W$ of $U$.
- It follows that $W$ is a neighborhood of $A.$
- Thus $W\in\mathcal N(A).$∎

**Grotemeyer, K.P.**: "Topologie", B.I.-Wissenschaftsverlag, 1969