Open in topological spaces and the neighbourhood property In metric spaces we know that every open set is one that has for each point a small neighbourhood that belongs to the set. But is this still true if we are talking about topological spaces? I am not sure about this. Is it sufficient in topological spaces to say that each point of a set has a small neighbourhood that is also part of the set in order to determine whether a given set is open or not?
 A: There are two definitions of a neighbourhood of $x$: 
(1) the "strict" one, where $N$ is a neighbourhood of $x$ when $N$ is open and $x \in N$.
(2) the "lax" one, where $N$ is a neighbourhood of $x$ when there exists an open set $O$ such that $x \in O \subset N$.
Personally, I find the second more useful (I can talk about closed neighbourhoods, compact neighbourhoods etc. without needing them to be open).
For both definitions the following is true for a topological space $(X, \mathcal{T})$:
$O$ is open (i.e. $O \in \mathcal{T}$) iff for every $x$ in $O$ there exists a neighbourhood $N_x$ of $x$ such that $N_x \subset O$.
Proof: if $O$ is open then for every $x \in O$ we can pick (for both definitions) $N_x = O$. This proves one implication. Suppose the right hand side holds. We can, for every $x \in O$, find open sets $O_x$ such that $x \in O_x \subset N_x$. (for definition (2) this is clear, for one even clearer as then we can take $N_x = O_x$ everywhere). But then $O = \cup_{x \in O} O_x$ (every $x$ in $O$ is in its own $O_x$, and $O_x \subset N_x \subset O$ for all $x$, so we get no more), which shows $O$ is a union of sets in the topology, so itself in the topology, i.e. open.
