Given the neighbourhood axiomatization of topological structure, how do you prove that every neighbourhood $N$ of $x$ contains an open set $U$ where $x \in U$?
In more detail:
Consider the Axioms for neighbourhoods from Section 2.1 of Ronald Brown's textbook Topology and Groupoids
The textbook is available for free online on Ronald Brown's website: http://groupoids.org.uk/topgpds.html.
The direct link to the pdf is: https://groupoids.org.uk/pdffiles/topgrpds-e.pdf.
The definition is given verbatim as
Let $X$ be a set and $\mathcal{N}$ a function assigning to each $x$ in $X$ a non-empty set $\mathcal{N(x)}$ of subsets of $X$. The elements of $\mathcal{N(x)}$ will be called neighbourhoods of $x$ with respect to $\mathcal N$ (or, simply, neighbourhoods of $x$). The function $\mathcal N$ is called a neighbourhood topology if Axioms N1-N4 below are satisfied; and then $X$ with $\mathcal N$ is called a topological space.
The following axioms must hold for each $x$ in $X$.
N1 If $N$ is a neighbourhood of $x$, then $x\in N$.
N2 If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$.
N3 The intersection of two neighbourhoods of $x$ is again a neighbourhood of $x$.
N4 Any neighbourhood N of $x$ contains a neighbourhood M of x such that $N$ is a neighbourhood of each point of M.
I think I have seen definitions where the fourth axiom above is stated as:
N4*: Any neighbourhood N of $x$ contains a neighbourhood M of x such that $M$ is a neighbourhood of each element in itself.
Or in other words, that any neighbourhood $N$ of $x$ contains an open set $M$ containing $x$ (where the definition of an open set in terms of neighbourhoods is a set $U$ which is a neighbourhood of all points in $U$). Now, I see that N1-N3+N4* implies N1-N4 because of N2. My question is how to prove N4* from N1-N4.
Even if I am misremembering and N4* is not used anywhere, I think it is still true because of the equivalent definition of a topological space in terms of open sets. In that approach, one recovers the neighbourhoods by defining that a set $N$ is a neighbourhood of $x$ if it contains an open set $U$ and $x \in U$.
I also read through some of Nicolas Bourbaki's text Elements of Mathematics General Topology and found the same set of axioms as N1-N4 (in Chapter 1 Topological structures in the subsection on Neighbourhoods). More specifically:
In that text, the definition of topological structure is presented in terms of open sets.
The text then moves on to define neighbourhoods
Definition 4. Let $X$ be a topological space and $A$ any subset of $X$. A neighbourhood of $A$ is any subset of $X$ which contains an open set containing $A$. The neighbourhoods of a subset $\{x\}$ consisting of a single point are also called neighbourhoods of the point $x$.
It then states
Proposition 1: A set is a neighbourhood of each of its points if and only if it is open.
After this, the properties of neighbourhoods of $x$ are listed as $\mathbf{V_I}$-$\mathbf{V_{IV}}$. The text says
Let us denote by $\mathcal{B(x)}$ the set of all neighbourhoods of $x$. The sets $\mathcal{B(x)}$ have the following properties:
($\mathbf{V_I}$) Every subset of $X$ which contains a set belonging to $\mathcal{B(x)}$ itself belongs to $\mathcal{B(x)}$.
($\mathbf{V_{II}}$) Every finite intersection of sets of $\mathcal{B(x)}$ belongs to $\mathcal{B(x)}$.
($\mathbf{V_{III}}$) The element $x$ is in every set of $\mathcal{B(x)}$.
($\mathbf{V_{IV}}$) If $V$ belongs to $\mathcal{B(x)}$, then there is a set $W$ belonging to $\mathcal{B(x)}$ such that, for each $y\in W$, $V$ belongs to $\mathcal{B(y)}$
However, the text then states that:
By Proposition 1, we may take $W$ to be any open set which contains $x$ and is contained in $V$.
But I do not see how one can take $W$ this way.
I have thought about this and searched for an answer but haven't had any luck. Am I missing something trivial? Any help is appreciated.
