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Something is bugging me about Mendelson's Introduction to Topology Chapter $3$ Definition $3.4.$

Quote from the book: "Definition $3.4:$ Let $X$ be a set. For each $x\in X,$ let there be given a collection $\mathfrak{N}_x$ of subsets of $X$ (called the neighbourhoods of $x$) satisfying the conditions $N1-N5$ of Theroem $3.1.$ This object is called a neighbourhood space."

Does it mean a neighbourhood space of a fixed $x,$ i.e., does it mean:

Definition $3.4:$ Let $X$ be a set. Let $x\in X.$ Let there be given a collection $\mathfrak{N}_x$ of subsets of $X$ (called the neighbourhoods of $x$) such that $\mathfrak{N}_x$ satisfy the conditions $N1-N5$ - with the words "For each point $x...$" omitted - of Theroem $3.1.$ This object is called a neighbourhood space (of $x$).

More context:

Quote from the book:

"Theorem $3.1\quad $ Let $(X, \mathfrak{I})$ be a topological space.

$N1.$ For each point $x\in X,\ $ there is at least one neighbourhood $N$ of $x.$

$N2.$ For each point $x\in X$ and each neighbourhood $N$ of $x,\ x\in N.$

$N3.$ For each point $x\in X$ if $N$ is a neighbourhood of $x$ and $N'\supset N,$ then $N'$ is a neighbourhood of $x.$

$N4.$ For each point $x\in X$ and each pair $N,M$ of neighbourhoods of $x,\ N\cap M$ is also a neighbourhood of $x.$

$N5.$ For each point $x\in X$ and each neighbourhood $N$ of $x,$ there exists a neighbourhood $O$ of $x$ such that $O\subset N$ and $O$ is a neighbourhood of each of its points. "

Now I have read further on and get the gist of what is going on, but now I am getting into the detail; hence my question. I read back over the chapter up until this point, but it didn't resolve my specific question.

My thought on why the answer to my question might be negative: Let $X=\mathbb{R}^2.$ Then define $\mathfrak{N}_x$ to be the set of all open balls in $\mathbb{R}^2.$ I am not sure if this is a neighbourhood space, but it could be if you interpret the nbhds of a specific $y\in \mathbb{R}^2$ as members of $\mathfrak{N}_x$ which contain $y.$ For example, $N2$ could be interpreted as true because for each $y\in X=\mathbb{R}^2,$ there is at least one nbhd of $y,$ i.e. member of $\mathfrak{N}_x$ which contains $y.$ However, for any $y\in X,$ there are members of $\mathfrak{N}_x$ that do not contain $y.$ So it is not clear to me what the intention is and if a nbhd space is meant to be a "nbhd space" in it's own right like my example here, or if a nbhd space is meant to be a "nbhd space of a specific $x\in X$", as I more clearly defined above. I think part of my confusion comes from the fact that Definition $3.4$ starts with "for each $x,$" and each "axiom" in Theorem $3.1$ starts with "For each point $x...$".

Adam Rubinson
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  • I see. $N5$ prevents my definition $3.4$ from making sense. So if let $X=\mathbb{R}^2,$ and for each $x\in X,$ let $U_x$ be the set of all subsets of $X=\mathbb{R}^2$ that contain an open ball that contains $x.$ (So for each $x,\ U_x$ is the set of all "neighbourhoods" of $x$ in the metric space sense - note that these nbhds are not necessarily connected - if they were, then $N3$ would fail). If we define $\mathfrak{I}:= \displaystyle\bigcup_{x\in\mathbb{R}^2} U_x,$ then the collection $\mathfrak{I}$ of subsets of $X$ is a neighbourhood space because $N1-N5$ are satisfied. Is this correct? – Adam Rubinson Oct 02 '24 at 12:35
  • I mean, he says "collection", not "family"... I feel like if he meant family he would have said family (as he does go into detail of what a family is in chapter $1).$ – Adam Rubinson Oct 02 '24 at 14:02
  • Also, when you say you "forget a lot of information", you could argue that you "get that information back again" when you re-visit axioms $N1-N5.$ (when trying to prove something is a neighbourhood space or not). But I do see what you mean. Anyway, we're quibbling about a minor thing now. Thanks for your help. – Adam Rubinson Oct 02 '24 at 14:10
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    I must insist. This is not minor thing. It is essential. It is absolutely impossible to "get that information back again", i.e. to guess what each $\mathfrak N_x$ was, if you only know their union – Anne Bauval Oct 02 '24 at 14:31

1 Answers1

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If you define the neighborhoods of $y$ as the open balls containing $y$, property N3 is not satisfied because a superset of an open ball is not necessarily an open ball.

More importantly, though beginning with "For each point $x$... " like N1 to N4, N5 refers to the neighborhoods of many other points than $x$. This non-locality prevents your own version of Definition 3.4, and the notion of "neighbourhood space of a fixed $x$", to make sense.

What Mendelson calls "a neighborhood space" is a set $X$, together with a family $(\mathfrak{N}_x)_{x\in X}$, each $\mathfrak{N}_x$ being a set (or, as he says, a "collection") of subsets of $X$, satisfying N1 to N4, and such that moreover, the family $(\mathfrak{N}_x)_{x\in X}$ satisfies N5.

Anne Bauval
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