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Let $X$ be a set of points, along with a certain function $N$ that assigns a subset to every $x \in X.$ Then $X$ together with $N$ define a topological space if $N$ satisfies the following axioms: (quoting below from wikipedia)

  1. If $N$ is a neighbourhood of $x$ (i.e., $N \in N(x)$), then $x \in N$. In other words, each point belongs to every one of its neighbourhoods.
  2. If $N$ is a subset of $X$ and contains a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. I.e., every superset of a neighbourhood of a point $x$ in $X$ is again a neighbourhood of $x$.
  3. The intersection of two neighbourhoods of $x$ is a neighbourhood of $x$.
  4. Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.

Questions:

  1. My first struggle is with the statement of the second axiom, namely that any superset $S\supseteq M$ of a neighbourhood $M$ of $x$ is again a neighbourhood of $x$.

    • a) Doesn't this mean that everything is a neighbourhood of everything because I can just set $S=X$? But then what good $N$ is at characterizing $X$ in terms of specific subsets of every $x$, if any neighbourhood can be indefinitely enlarged?
    • b) How to understand the need for such an axiom?

  2. Am I right to interpret the 4th axiom as follows: Given a neighbourhood $N$ of $x,$ it is always possible to extract a smaller neighbourhood $M'\subseteq M$ from $M,$ such that $M'$ is an open subset, where $M$ need not be open itself.

  3. To me it seems that the 2nd and 4th axioms are equivalent statements. What makes them distinct? (or in other words what makes the 4th axiom necessary).

Andrés E. Caicedo
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    It is true that any neighborhood can be indefinitely enlarged. But it cannot arbitrarily reduced. It can perhaps be indefinitely reduced, in the sense that we can always find a smaller neighborhood within a neighborhood, but not in the sense that any subset containing $x$ of the given neighborhood $M$ of $x$ is a neighborhood of $x$. This is opposed to axiom 2 which means that indeed any superset of $M$ is again a neighborhood of $x$, especially $X$. – Stefan Hamcke Oct 29 '15 at 13:02
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    It is true that any neighborhood $M$ of $x$ contains an open neighborhood of $x$, but this requires that you have defined what "open" means. The characterisation by neighborhoods is actually equivalent to that by open sets, but this requires us to use a suitable definition of open sets in terms of neighborhoods (and, if we start with open sets, we need to define neighborhoods in terms of open sets in an appropriate manner). – Stefan Hamcke Oct 29 '15 at 13:07
  • It's good to think of an example. In the standard topology on $\Bbb R$, a set $N$ is a neighborhood of $x$ if there are reals $a$ and $b$ such that $x\in(a,b)\subseteq N$. – Akiva Weinberger Oct 29 '15 at 13:41
  • @StefanHamcke Dear Stefan, very interesting. Would you mind writing this up as an answer? (with potentially more details if you find the time). –  Oct 30 '15 at 14:04

1 Answers1

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The enlargement axiom does not trivialise the theory. It does imply that $X$ is a neighbourhood of every $x \in X$: every $x \in X$ has some $N \in N(x)$ (this is actually part of the axioms/conditions, often forgotten: every $N(x)$ is non-empty) and $N \subseteq X$ so $X \in N(x)$ by axiom 2. And everything inbetween $N$ and $X$ is a neighbourhood too. We cannot go smaller in general, so it's not true that everything becomes a neighbourhood of everything: $X$ is a neighbourhood of every point, but any fixed set $A$ is only a neighbourhood of $x$ if it happens to be the case that $A \in N(x)$, and otherwise it's not.

As to the necessity of the axiom: look at metric spaces, where $A$ is a neighbourhood of $x$ iff there is some $r > 0$ such that $B(x,r) \subseteq A$. The way this is defined already implies that larger sets than $A$ are also neighbourhoods of $x$; the same $r$ will work. So it's not a very surprising axiom. The same can be said for ordered spaces ($A$ is a neighbourhood of $x$iff either there exists $a \in X$ with $x < a$ and $\{z : z < a\} \subseteq A$, or there exists $b \in X$ with $b < x$ and $\{z : b < z \} \subseteq A$ or there exist $a,b \in X$ with $a < x, x < b$ and $\{z: a < z, z < b \} \subseteq A$), where also the enlargement is clearly true. We really are interested in the "small" neighbourhoods of $x$ (for continuity etc.) but in this axiomatisation we want to capture the notion of all possible neighbourhoods, even the not so interesting large ones. We could also axiomatise "basic" neighbourhoods instead, instead of all neighbourhoods, but this is not what we are doing here. In short, the relation of $A$ being a neighbourhood of $x$ is "local" and not affected by enlarging the set.

As to axiom 4, it is true that the "intended" result is that every neighbourhood of $x$ contains an open neighbourhood of $x$, but for that we first need to define what "open" means. We only have a set of sets $\{N(x): x \in X\}$ satisfying the axioms, no notion of openness. But if you already know the standard axioms for topology in terms of open sets, you want to go back and forth:

If you have a topology $\mathcal{T}$, define for each $x \in X$: $N(x) = \{A \subseteq X: \exists O \in \mathcal{T}: x \in O \subseteq A\}$ and check that this fulfills the axioms (again see that 2 is trivially fulfilled). Call this system $\mathcal{N}(\mathcal{T})$. On the other hand, if we have a system $\mathcal{N}= \{N(x): x \in X \}$ satisfying the axioms, define $\mathcal{T} = \{O \subseteq X: \forall x \in O: O \in N(x) \}$ and check that this defines a topology (we do not need axiom 4 for this, it turns out). Call this topology $\mathcal{T}(\mathcal{N})$. The axioms 4 is used to show that $\mathcal{N}(\mathcal{T}(\mathcal{N})) = \mathcal{N}$ again. So the neighbourhood system from the topology induced by the neighbourhood system is that same one we started with. It's a fun exercise to check this out in detail. It will enhance your understanding of these axioms. It's also true that $\mathcal{T}(\mathcal{N}(\mathcal{T})) = \mathcal{T}$, which is also fun to prove.

Henno Brandsma
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  • Thanks for your explanations, I need some time to digest some key parts. But if I may already ask some questions: i) why can't I interpret from your first paragraph that: since $X \in N(x)$ then for any subset $A \subseteq X$ by implication $A \subseteq N(x)$ thus concluding that any subset $A$ is a neighbourhood of $x$. ii) I m a bit lost about the openness not being defined, I mean the axiom 4 is making an statement about sth undefined? How many ways are there to define openness? Sorry for these very naive questions :( –  Oct 29 '15 at 15:49
  • The axioms says that a superset of a neighbourhood is one too. You go back ($A \subseteq X, X \in N(x)$ implies $A \in N(x)$?) which is not allowed by that axiom. $A \subseteq N(x)$ makes no sense, set theory wise.. – Henno Brandsma Oct 29 '15 at 16:06
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    Axiom 4 says nothing about openness in itself. Purely logically it says: $\forall x \in X \forall N \in N(x) \exists M \in N(x) \forall y \in M: N \in N(y)$. It's purely defined in terms of the families $N(x)$. Openness can then be defined in terms of neighbourhoods (or vice versa). The point is showing they come down to the same notion (so the different axioms can be used to do eventually the same things). – Henno Brandsma Oct 29 '15 at 16:10
  • Thanks again. I will think about all this, and get back to you. –  Oct 29 '15 at 17:41
  • BTW is there any specific book that you would point a beginner to for getting acquainted with basics of topological spaces (such as the discussions above)? (do you know Introduction to Topology by Mendelson?) –  Oct 30 '15 at 13:48
  • No, I don't know Mendelson. I think Munkres has some discussion, and Engelking (general topology) in its first chapter (but goes beyond beginner level quite quickly). – Henno Brandsma Oct 30 '15 at 13:56
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    Dear Brandsma, I have read your answer carefully, I understand a lot better now, specially your second paragraph. One question regarding notation: Am I correct in interpreting ${z : z<a}$ as "There exists $z$'s such that they are smaller than $a$"? One question about the last paragraph: so basically either we start from neighbourhoods and use the 4th axiom to invoke openness to define a topology $\mathcal{T}$ or alternatively, we start from a topgoloy $\mathcal{T}$ and use axiom 2 to show that for every open subset in $\mathcal{T}$ there exists a superset, being a neighbourhood $N \in N(x)$ –  Oct 30 '15 at 14:30