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I am looking at the section "Motivation" for the Wikipedia entry on "open sets": and I am not sure it is doing such a good object of motivating open sets, as opposed to closed sets.

I quote: "Intuitively, an open set provides a method to distinguish two points. For example, if about one point in a topological space there exists an open set not containing another (distinct) point, the two points are referred to as topologically distinguishable. In this manner, one may speak of whether two subsets of a topological space are "near" without concretely defining a metric on the topological space. Therefore, topological spaces may be seen as a generalization of metric spaces."

It seems to me (I am sure this is too naive) that I could equally well distinguish two points by using closed sets instead of opens. What am I missing? What is so crucial about opens that closed sets don't have/do?

Update: I get the feeling that maybe opens are "less precise" than closed sets, in the sense that the boundary of a set, that the closed sets have, is an object with a rather precise "position". Opens seem to avoid having to be that precise about anything - I know this is vague. It might even be wrong, but it seems like it could be a useful intuition to differentiate opens and closeds? Or not?

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See this:… – Michael Greinecker Aug 12 '13 at 21:47

You could just as well define your topology by defining the closed sets, demanding that your space and the empty set be closed, and that arbitrary intersections and finite unions of closed sets are closed. Since open sets are the complements of closed sets, this would then give us the open sets, and define the exact same topology.

For a more pedagogic or philosophical answer, see this mathoverflow thread for some good discussion on the motivation/interpretation of open sets:

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Since closed sets are, by definition, exactly those whose complements are open, we could do everything just as well by speaking about closed sets instead of open ones. The fact that open sets are the most common way to define a topology is purely conventional.

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OK - so when opens are used in talking about continuous functions, I can replace them with closed sets and not lose anything? – Frank Aug 12 '13 at 19:49
@Frank: Not word for word, of course, but you can reword the definitions to have the same content but uses "closed" as a primitive concept rather than "open". – Henning Makholm Aug 12 '13 at 19:52
But then - why do we bother distinguishing these 2 kinds of sets at all? I mean, is there a point/a theorem... that works for opens but not for closeds? This distinction must be more than pure "botany"? – Frank Aug 12 '13 at 19:58
You have to distinguish them, because they have different properties. The properties are in some sense complementary, since closed sets are the (set-wise) complements of open sets, but they're still different. For example, arbitrary unions of open sets are open, but infinite intersections need not be. For closed sets, the opposite is true: arbitrary intersections of closed sets are closed, but infinite unions need not be. This duality is because a union of closed sets is a union of complement of open sets, which is the complement of an intersection of open sets. – Donkey_2009 Aug 12 '13 at 20:50
@Kaz: You're misunderstanding. It's not that an open set and a closed set differ in this respect, it's that the properties of openness and closedness differ in this respect. More explicitly . . . let the property "is an open set" be denoted O, and the property "is a closed set" be denoted C. Then O is closed under union and under finite intersection (but not arbitrary intersection), and C is closed under intersection and under finite union (but not arbitrary union). So these properties have different properties; they are distinguishable. – ruakh Aug 13 '13 at 0:38

Your intuition about the "fuzziness" of open sets is correct.

Inuitively, if $U$ is an open set in the topological space $X$ and $x \in U$, then $U$ necessarily contains any other point in $X$ that is "near enough" to $x$.

Now this statement has a precise meaning in a metric space: in that setting, "near enough" means "within distance $\epsilon$" for some $\epsilon > 0$", and then the preceding statement becomes the definition of an open subset of a metric space. In a general topological space, open sets are what provide the sense of "nearness", so the preceding statement is circular at best, but it is still what you should keep in mind. You should contrast it with closed sets: if $x$ is a point on the boundary of a closed set $F$, then there will be points in $X$ that come as close as you like to $x$ but are not in $F$ (the points on the "other side of the boundary" to $F$, intuitively speaking: think of the end-points of a closed interval, or a point on the boundary of a closed disk).

The preceding discussion is purely intuitive: to really see how the definitions of closed and open sets are used you have to get a little further in your study of topology and see the technical way in which these notions are used. If you haven't already done so, it can be good to start with the metric space case: then the open and closed subsets are defined in terms of the metric, as is the concept "sufficiently close to $x$", and the above intuitions become precise.
Then, in your study, you will see that lots of concepts can be defined, and that lots of theorems can be proved, purely in terms of open and closed subsets, without having to explicitly refer to the metric. (E.g. the definitions and basic properties of continuity, convergence, compactness, connectedness, etc.) At this point it will seem natural to abstract out the properties of open and closed sets, without using the crutch of a metric to define them.

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Hmmm... what I had in mind is that in some sense, "closed" implies some sort of "discontinuity". Let's not involve metric spaces, but with a closed set, there are things happening "on one side" of a point, that are not happening "on the other side". Not so with opens: whatever happens "on one side", happens "all around" the point - no "discontinuity", no border is reached. I had something like that in mind, however non-mathematical it is :-) – Frank Aug 12 '13 at 21:58
@Frank: Dear Frank, That's a fairly good intuition. Regards, – Matt E Aug 12 '13 at 22:12

One way that open sets are more useful than closed sets is the following fact:

If $(X,\tau)$ is a topological set, and $U$ is a subset of $X$, then $U$ is open if and only if it has the property that for every $x\in X$, there is an open set $V_x$ satisfying $x\in V_x\subset U$ (We call $V_x$ an open neighbourhood of $x$).

The proof is very easy: of course, if $U$ is open, we can just take $V_x=U$ for each $x$. Conversely, note that:

$$ U \subset \bigcup_{x\in X}V_x\subset U $$

So $U=\bigcup_{x\in X}V_x$, which is the union of open sets and therefore open.

If you replace 'open' with 'closed' in the above fact, the result is not true. For example, every point in the open interval $(0,1)$ has a closed neighbourhood in that set, yet $(0,1)$ is not closed. The proof given for open sets above fails because infinite unions of closed sets need not be closed.

I like this fact, because it's very similar to the definition of open sets in metric spaces: a set is open if every point in the set has an open ball around it that is contained in the set. Obviously, you can't use the fact above as a definition of open sets, since it involves open sets, but the same thing is going on. Very often in topology you define a basis of open sets - some collection of open sets rather like the open balls in metric spaces. Other open sets are then sets in which every point has a neighbourhood that is one of the basic open sets.

Intriguingly, we can change the definition of an open set to the following: a set $U\subset X$ is open if and only if for every $x\in U$ there exists $\varepsilon_x$ such that the closed ball $B_x(\varepsilon_x)$ is contained in $U$. It's fairly easy to see that this definition is exactly equivalent to the open-balls definition: every open ball contains a closed ball, and every closed ball contains an open ball. But why would you want to define open sets using closed balls? It makes much more sense to define them using open balls.

In the end, it turns out to be much more useful to have arbitrary unions of open sets being open than it does to have arbitrary intersections of closed sets being closed. That said, a lot of topological proofs are more naturally stated using closed sets (such as my favourite proof of the de Bruijn-Erdős theorem or Furstenberg's celebrated proof of the infinitude of primes). In addition, some topologies, such as the Zariski topology, are more accurately defined by specifying the closed, rather than the open sets.

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My way of looking at this is simply as a generalisation. Take the definition of continuity as it is in real analysis: $$\forall \varepsilon > 0\ \ \exists \delta > 0 : |x-c|<\delta \implies |f(x)-f(c)|<\varepsilon $$

And now we can use this to hopefully see what is important, and use it to generalise our notion of continuity to apply it to more general contexts. The approach metric spaces have is to notice that the defintion says something about the distances between $x,c$ and $f(x),f(c)$, and hence defines continuity in terms of whatever is chosen to define 'distance' on a more general space.

Topology on the other hand, notices that the definition sends the open set $|x-c|<\delta$ to the open set $|f(x)-f(c)|<\varepsilon$. From this, topology preserves this idea (which can be proven in the context of analysis; that open images have open preimages iff the map is continuous) and uses it to define continuity.

This, to me at least, is why metric spaces deal with distances, and topological spaces open sets.

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Note that the definition of continuity for a function between general topological space is NOT that it sends open sets to open sets, but that the preimage of an open set is open. – Devlin Mallory Aug 12 '13 at 21:00
What happens if we replace $< \epsilon$ by $<= \epsilon$ in the definition of continuity? If it still works, there are two possible generalizations, one with opens and one with closeds, and my question is really: why do we prefer one over the other? If it doesn't work, then opens are required, because they do something that closeds don't do. One could also ask what happens if the definition holds for some $c$ with $< \epsilon$ and for some $c$ with $<= \epsilon$. Does it matter? – Frank Aug 12 '13 at 21:25
@Frank: Dear Frank, You are right that we can replace $< \epsilon$ by $\leq \epsilon$ in the definition of continuity, but a typical closed set does not contain closed balls of radius $\epsilon$ for some $\epsilon \geq 0$ around all its points. So despite your observation, the properties of closed and open sets are not the same. See Donkey_2009's answer for more on this. Regards, – Matt E Aug 12 '13 at 21:43
@Matt - not only are they not the same, but opens are actually much "nicer" because they don't imply this "discontinuity" on at least one side of some points - by the way for continuity around a point c, you would want your set to be open on "all sides" of c, otherwise, wouldn't you possibly get a "discontinuity" on some side? – Frank Aug 12 '13 at 22:57

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