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My question concerns the Definition via Neighborhoods of a topological space. I am having trouble seeing why $(4)$ is necessary in the definition, which I'll briefly run through below.

Let $X$ be a set and define a function $\mathcal{N}:X \to \mathcal{P}(X)$ which assigns to each $x \in X$ a nonempty collection of subsets of $X$, which we call neighborhoods of $x$, satisfying the following properties:

  1. If $N \in \mathcal{N}(x)$ then $x \in N$.
  2. If $N,M \in \mathcal{N}(x)$ then $N\cap M \in \mathcal{N}(x)$.
  3. If $N \in \mathcal{N}(x)$ then any set $Y\supseteq N$ is also in $\mathcal{N}(x)$.
  4. If $N \in \mathcal{N}(x)$ then there exists an $M \subseteq N$ in $\mathcal{N}(x)$ such that $N \in \mathcal{N}(y)$ for all $y \in M$.

Define a set $U \subseteq X$ to be open in $X$ if $U$ is a neighborhood of all of its points.

As I mentioned above, I am not sure why we need $(4)$ because (I think) I can recover the standard definition of a topology via open sets just from $(1)$ through $(3)$. Assuming the definition of an open set given above we see that clearly $\varnothing$ is open because it is vacuously a neighborhood of all of its points. Moreover, $X$ is open by $(3)$. Finite intersections of open sets are open by $(2)$, and arbitrary unions of open sets are open by $(3)$.

Wikipedia mentions that $(4)$ "has a very important use in the structure of the theory", but I don't see how this is possible if I can recover the standard definition of a topology via open sets without it.

I feel that I must be missing something simple. Have I made a mistake? Do you actually need $(4)$ to recover the standard definition of a topology via open sets? If not, then how is $(4)$ important?

wgrenard
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    It is there such that $\mathcal{N}$ is a system of neighborhoods for the corresponding topology. It is true that the sets defined as open above would form a topology, but without (4) some elements of $\mathcal{N}(x)$ might not be actual neighborhoods of $x$ (sets containing an open set containing $x$) –  Mar 15 '18 at 18:44
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    Take into account that (4) is the only property that relates neighborhoods of a points with neighborhoods of another. Without it you can put in $\mathcal{N}(x)$, for some $x$ more sets, without having any effect in the corresponding topology. For example, take the indiscrete topology in $\mathbb{R}$. Define $\mathcal{N}(x)=\mathbb{R}$ for all $x\neq0$. No matter how you define $\mathcal{N}(0)$ satisfying (1)-(3), the topology that you would get by the open sets above is the same, the indiscrete topology. –  Mar 15 '18 at 18:56
  • @AOlov, you should turn those two comments into an answer! – Mees de Vries Mar 15 '18 at 18:57
  • @AOlov Thank you, that's what I was missing. If you don't mind me asking a question to clarify . . . Based off of what you're saying I would want $(4)$ to say, "If $N \in \mathcal{N}(x)$ then it should contain an open neighborhood containing $x$. (i.e. it should contain an $M \in \mathcal{N}(x)$ which is a neighborhood of all of its points)." What it actually says is that $N$ should contain an $M \in \mathcal{N}(x)$ such that $N$ is a neighborhood of all of the points in $M$. Are these significantly different? – wgrenard Mar 15 '18 at 19:19
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    You can prove it as an exercise. Given $A\in\mathcal{N}(x)$ look at $A^o={y\in A:\ A\in\mathcal{N}(y)}$. Take $y\in A^o$. Then $A\in\mathcal{N}(y)$. By (4) there is $\mathcal{N}(y)\ni M\ni y$ such that $A\in\mathcal{N}(z)$ for all $z\in M$. Therefore $M\subset A^o$. By (3) $A^o\in\mathcal{N}(y)$. Therefore, $A^o\subset A$ is open. –  Mar 15 '18 at 20:06
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    It is a good exercise to do once in live (but probably no more than once) to prove the equivalence of the ways to axiomatize topology: Open sets, closed set, neighborhoods, Kuratowski operator (closure operator), interior operator. –  Mar 15 '18 at 20:11
  • Similarly, in the Kuratowski closure operator definition, it suffices to take a preclosure operator to define a topology (which is a closure operator minus the axiom $\operatorname{cl}(\operatorname{cl}(X)) = \operatorname{cl}(X)$. What doesn't happen in this case, in general, is that the closure operator induced by this topology recovers the preclosure operator that we started with. – Daniel Schepler Mar 15 '18 at 22:12
  • In addition to the definitions @AOlov mentioned, there are also the internal conditions for a collection of subsets to form an open basis of some topological space; and similarly, the internal conditions for a collection of subsets indexed by $X$ to form a system of neighborhood bases of some topological space. The latter is arguably the "natural" way to define a topology from a metric, for example. – Daniel Schepler Mar 15 '18 at 22:25
  • In axiom 4, shouldn't $M$ be nonempty? Otherwise it seems like $M=\varnothing$ is always satisfied and the axiom doesn't seem to add any new content. – user326210 Mar 15 '18 at 22:35
  • Thanks all for your input. I was stuck on making sure that I could define open sets correctly and forgot all about looking at whether or not neighborhoods defined in one perspective were the same as neighborhoods in the other perspective. @user326210 Yes it should be but since $M \in \mathcal{N}(x)$ it must contain at least $x$. – wgrenard Mar 15 '18 at 22:40
  • @wgrenard Oh I see. Thanks! – user326210 Mar 15 '18 at 22:41

2 Answers2

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  1. When you define topological terms using open sets as a foundation, open sets are defined axiomatically and neighborhoods are defined as any set containing an open set.

  2. When you define topological terms using neighborhoods as a foundation, neighborhoods are defined axiomatically and open sets are defined as any set which is a neighborhood of all of its points.

  3. These definitions coincide if you use the four axioms above. As you point out, the first three are enough to make sure that open set means the same thing in both scenarios. The fourth axiom is necessary to make sure that neighborhood means the same thing.

  4. If you exclude the fourth axiom, then your definition might allow a neighborhood (axiomatically defined) which is not a neighborhood (a set containing an open set).

    For example, take the real line $\mathbb{R}$ and define a neighborhood system axiomatically as follows. For each point $x\neq 0$, let $\mathbb{R}$ be its only neighborhood. Let the interval $(-1,1)$ and its supersets be the neighborhoods of $x=0$. You can show that the only open sets are $\mathbb{R}$ and $\varnothing$ [*proof below]. It follows that $(-1, 1)$ is a neighborhood of 0 according to neighborhood axioms, but not a neighborhood of 0 according to open set axioms— it contains only the trivial open set $\varnothing$.

    [*] Proof of claim: if $U$ is an open set and $U$ contains $x\neq 0$, then $U=\mathbb{R}%$ because $\mathbb{R}$ is the only neighborhood of $x$. If $U$ contains only 0, then $U=\{0\}$, which isn't an open set because it isn't a neighborhood. If $U$ is empty, then $U$ is trivially an open set.

user326210
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You need it because the intended meaning for neighbourhood of $x$ is that it contains an open set containing $x$. Also, we want $O$ open to mean "$O$ is a neighbourhood of each of its points". Axiom (4) ties this together, and also guarantees that when we define a topology from the neighbourhood system, that the neighbourhoods for that newly defined topology are exactly the neighbourhoods we started out with; it gives a bijection between neighbourhood systems and topologies.

Henno Brandsma
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