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I'm studying topology since a few months ago and I have never caught a good intuition of the topological spaces, but now I think that I did.

My intuition is the next; as many people point out the open sets capture a notion of nearness, (but I did never understand the exact way of this nearness intuition),the points in the open sets are near, and two points in distinct open sets are far, and then the axioms try to capture the behaviours of the opens.

To find a good intuition I have worked with two examples of spaces which I named $A$ and $B$; $A=[\{a,b,c,d,e,f\},(\{a,b,c,d,e,f\},\{\},\{a,b,c,e,f\},\{e,f\},\{a,b\},\{a,b,d,e,f\},\{a,b,e,f\})]$, $B=[\{a,b,c,d,e,f\},(\{a,b,c,d,e,f\},\{\},\{a,b,c,e,f\},\{a,b,d,e,f\},\{a,b,e,f\})]$
and I obtained the next pictures (and here is my first question: are these the correct spatial representaition of the spaces? $A$ is in the right, $B$ is the left, the black circles represent points):

enter image description here

And then the union axiom allows to establish different degrees of nearness, specifically the opens that aren't union of other opens have near points, and the unions have points nearer than other points that aren't. Concretely in the space $A$; $\{e,f\}$ and $\{a,b\}$ are nearer than $c$ and $d$, then the unions allow to speak of a "global " nearness.

The intersection axiom also talk about different degrees of nearness, in the way that if we have near points and we have other near points that have common points, then than points are near, and in fact they are nearer, and so the intersection talk of a "local" nearness.

And finally the axiom of the necessary membership of the space itself and the empty set, guarantees the fact that "there are nothing outside the space" or the totality of the space and in that sense all the points are near in the space.

Is this a good intuition? if not, please give me a good one.

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    $\begingroup$ Why is it important to have metric-free intuition? Frankly, metric spaces are where almost everyone first gets intuition about topological spaces. I first read about topological spaces before seeing metric spaces and the concept made absolutely no sense. After I saw metric spaces, though, the transition to topological spaces was not that painful (just having to be a bit careful here and there). $\endgroup$
    – KCd
    Jan 18, 2013 at 5:04
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    $\begingroup$ Maybe you need a significant example of a topology that doesn't come from a metric. The example in your question is kind of fake; nobody cares about such a construction. Read about the Zariski topology, which is an important example of a nonmetrizable (even non-Hausdorff) topology. $\endgroup$
    – KCd
    Jan 18, 2013 at 5:06
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    $\begingroup$ You say you want a metric-free intuition, but I think this is a bad idea, because the axioms for a topological space are an abstraction of the behavior of open subsets of $R^n$. The original purpose of topology is to examine properties of $R^n$ in a more general setting. $\endgroup$
    – MJD
    Jan 18, 2013 at 5:12
  • $\begingroup$ Yes, but those spaces are not very intuitive! $\endgroup$
    – MJD
    Jan 18, 2013 at 5:18
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    $\begingroup$ See math.stackexchange.com/questions/31859/… . $\endgroup$ Jan 18, 2013 at 6:04

3 Answers 3

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Your desire for a metric-free intuition is, in some sense, futile due to the fact that every topological space $(X,\tau )$ is metrizable as long as by metrizable you mean the existence of a value quantale $V$ (a certain axiomatization of some of the properties of $\mathbb R$) together with a function $d:X\times X\to V$ satisfying $d(x,x)=0$ and $d(x,z)\le d(x,y)+d(y,z)$, such that the open-ball topology for this metric is the original topology.

This is explained in detail in all topologies come from generalized metrics by Ralph Kopperman and Quantales and continuity spaces by Robbert Flagg.

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  • $\begingroup$ Thanks! this is a very useful answer as well as interesting. $\endgroup$
    – Fenrir
    Jan 18, 2013 at 8:46
  • $\begingroup$ you're very welcome. Developing the right topological intuition takes time. The articles I mention frame things in a helpful way I find. $\endgroup$ Jan 18, 2013 at 8:49
  • $\begingroup$ Excellent answer. Especially for the links. $\endgroup$
    – user170039
    Jan 21, 2018 at 4:38
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    $\begingroup$ Yep, which moreover essentially means a topological space is what happens when you simply forget about exactly how "big" things are, and rather simply leave only the relationships between them formed by their fact of being bigger or smaller (or neither). $\endgroup$ Sep 14, 2019 at 12:05
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You’re much better off if you start with a known topological space, I suggest the real number system $\mathbb R$, and see how the axioms apply to it. Remember that a set $U$ of real numbers is open if $\forall x\in U$ there is $\varepsilon>0$ such that all the numbers within a distance of $\varepsilon$ from $x$ are in $U$. So your first task is to verify that an “open interval” such as $\langle0,1\rangle$ (the set of numbers strictly between $0$ and $1$) is open according to this definition. That is the beginning of your intuition. Now show that the “closed interval” $[0,1]$, namely the set of numbers between $0$ and $1$ including the endpoints, is not open. Now show that the intersection of two open sets is open. Now show that any union of open sets is open. Now show that the empty set is open. You have already shown that the whole set $\mathbb R$ is open, do you see why? Do you see why $0$ is “close” to $\langle0,1\rangle$ without being in this open set?

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  • $\begingroup$ thanks but I want a metric-free intuition. $\endgroup$
    – Fenrir
    Jan 18, 2013 at 4:49
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    $\begingroup$ It’s a benefit to mathematics to have people with different intuitions. So I applaud your desire. Yet, for most of us, our intuition comes from what we see in the world about us. And what we see is unquestionably metric. $\endgroup$
    – Lubin
    Jan 18, 2013 at 15:52
  • $\begingroup$ Could you please expand on "Do you see why 0 is “close” to ⟨0,1⟩ without being in this open set?". Thanks! $\endgroup$ Jul 20, 2014 at 11:00
  • $\begingroup$ Just because every neighborhood of $0$, and in particular each neighborhood $\langle-\varepsilon,\varepsilon\rangle$ has nonempty intersection with that open set. $\endgroup$
    – Lubin
    Jul 20, 2014 at 22:33
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You say that you don't want a "metric space" intuition, yet you want spaces to be "correctly spatially represented." I hate to say this, but all the phrases like "correctly spatially represented" and "nearby" are ideas of metric spaces. (For instance, you draw points close together in your pictures -- you're using the metric on your sheet of paper.) You won't get your non-metric intuitions using such thinking. So if you really demand some intuition that doesn't come from metric spaces, I suggest not using words like "nearness."

One intuition for open sets is that they are "fat" or "thick;" in fact, the "fattest" kinds of sets you can take in your given topological space.

A non-metric example of this intuition comes from the Zariski topology. If you don't know what this is, here is an illustration. Let $\mathbb C^n$ be the $n$-dimensional complex affine space. In the Zariski topology, a closed subset is a subset that is cut out by a collection of complex polynomials in the $n$ variables $z_1,\ldots,z_n$. For example, any singleton point is a closed subset (given by some equation $z_i = c_i$, where $c_i$ are the coordintes of the point) as are hypersurfaces given by some polynomial equation $f(\vec z) = 0$.

Note that the open ball (of say, radius $r < \infty$) in $\mathbb C^n$ is not an open set. (Its complement may be cut out by an inequality, but never by a complex polynomial.) In fact, this topology does not arise by putting a metric on $\mathbb C^n$. (And cannot; it's even non-Hausdorff.)

But its open sets are still "fat," in that open sets are always either empty, or complements of higher codimension subsets, so they are always top-dimension (i.e., n-complex dimensional) subsets. (By the way, if you are not comfortable thinking of the empty set as "fat," that's fine. All other open sets are "fat.")

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