1

Concerning open sets: A set $S$ is "open" if and only if it is a neighborhood of each of its points. But for $S$ to be a neighbourhood of its points if there is some other set $V$ which contains an open set $U$ which contains $S$. But We seem to get back then to that $U$ is open if there is some set $Z$ which contains an open set $Y$ containing $U$.

Eventually we run through all the letters of the alphabet and go on to funny symbols - depending of course on how large our original set $X$ is. However I find that in the definitions of neighbourhood system, $X$ is assumed to be a Topological space. A topological space however is such that it follows axioms concerning unions and intersections of open and closed sets - closed sets being defined if the compliments are open! It seems, at least to me, that there is some ambiguity somewhere. Could anyone elaborate on how all of this fits together?

Thanks,

Brian

Relative0
  • 1,449

3 Answers3

3

The terms "open set", "closed set", "neighbourhood", "closure", "interior" and so on are all interwoven and it is possible to pick any of them to define what a topology is. The common way is to use "open" as basic property (with the usual axioms, $\emptyset$, $X$, finite intersections, arbitrary unions) and define the other stuff in terms of it. But one could just as well start from the notion of closed sets (with different axioms, namely $\emptyset$, $X$, finite unions, arbitrary intersections; then define a set to be open if its complement is closed etc.) and define the rest from it. The same is possible with neighbourhood as fundamental property, or with closure as a fundamental operator, or ...

The notions by themselves however cannot be used to show that any set considered is open. For example one cannot show that $(0,1)\cup (2,3)$ is open because for each $x\in (0,1)\cup (2,3)$ some set $(x-\epsilon,x+\epsilon)$ is open and that this again is open because some smaller sets are open. No (and in fact $(0,1)\cup (2,3)$ is not an open subset of $\mathbb R$ if we chose a weird enough topology on $\mathbb R$). We must start with a given topology, i.e. a certain collection of subsets that obey the axioms described above. In case of $\mathbb R$, this topology is given by the fact that we want (among others) all open sets of the form $\{\,x\in\mathbb R: |x-a|<r\,\}$ to be open (topology induced by the metric) or because we want all sets of the form $(a,\infty)$ and all $(-\infty,a)$ to be open (topology induced by the order).

Nevertheless, it is a correct statement that a set is open if and only it is a neighbourhood for all its points. It's just that there is no need to recourse into any deeper levels as you supposed, just apply the definitions of one in terms of the other (or in terms of a third concept, see above).

Stefan Hamcke
  • 28,621
Hagen von Eitzen
  • 382,479
2

It's true that a set is open if and only if it's a neighborhood of each of its points, but this is not the definition! The definition is that a set is open if and only if it's an element of the topology, which can be any collection of subsets of $X$ closed under union and finite intersection (including the union and intersection of zero sets.)

Kevin Carlson
  • 54,515
  • if I understand what your saying correctly - this is the point I was trying to make, that it seems we can't just take some arbitrary family of sets $\mathcal{F}$ of subsets of some set $X$ and call some elements of $\mathcal{F}$ open as "open", seems to be defined in terms of a Topology. A topology however is defined (in one definition) by open sets - and in the definition concerning closed sets, they too are related to open sets via compliments. So I am stuck on what seems to me to be some sort of "circular" argument. – Relative0 Oct 25 '14 at 21:16
  • There's nothing circular. Open sets are defined as the elements of an arbitrary family of subsets of $X$ which obeys the axioms of a topology. I think your confusion is that the word "open" is usually used in a definition of topology. Instead, define a topology as a subset of the powerset of $X$ which contains the union of any set of its elements and the intersection of any finite set of its elements-and note that nowhere does this use the word "open!" Then, only after you've defined a topology, define a set to be open if it's an element of a topology. – Kevin Carlson Oct 26 '14 at 23:03
2

Usually, you start with a topology, i.e. $X$ with a collection $\mathcal{T}$ of subsets of $X$, that satisfies the usual axioms (closed under finite intersections, and arbitary unions). These sets are called open by definition.

Then you define a set $N$ to be a neighbourhood of $x$ when there exists an open set $O$ (So one from the topology $\mathcal{T}$) such that $x \in O \subset N$.

Then we get a theorem that $O$ is open iff it is a neighbourhood of each of its points.

On the other hand, a common path can also go as follows: we start with a metric space (say) (so $(X,d)$; usual axioms for a metric) and define a $N$ to be a neighbourhood of $x$ when there is some $r>0$ such that $B(x,r) \subset N$. Then we define an open set (in the topology derived from the metric) a set that is a neighbourhood of each of its points, in this sense.

Then we get a theorem that the set of all open sets (in the latter definition) is indeed a topology (so obeys those axioms). And the two notions of "being a neighbourhood of" then coincide again.

So the statement is not weirdly recursive at all, but it depends on what your starting point of definitions is.

Henno Brandsma
  • 250,824