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I've been studying introductory topology for a little bit now. I came across this video which explains open sets in a way I have never thought of. Even though the video is pretty elementary, I didn't know that open sets worked in this sense. If I were to 'imagine' moving toward the boundary of an open subset, I would just appear on the other side? Is this correct?

Is there any kind of intuitive way of thinking about the concept of an open set versus a closed set without just saying one is outlined with a dotted line versus a solid line. I ask this because I have never really understood why something like a domain is generally defined to be a set that is both open and connected. Why does it matter if it is open or closed? Is this just a formality? Or are open sets something that are special. I am curious because I would like to know if I have been thinking about open sets completely wrong.

Thanks.

Asaf Karagila
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Rellek
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    My answer to this question; http://math.stackexchange.com/questions/1185719/how-would-i-explain-an-open-set/1185749#1185749 might help for understanding an open set. – TravisJ Apr 24 '15 at 20:50
  • Yes. I don't know why this seems to mess with my head so much, it seems like there would always be some point right there before the boundary (ie dotted line) that would always have a neighborhood with points not belonging to the set ... but that would make it a closed set. After seeing the video, though, it seems as though every n-dimensional euclidean space can be transformed into an n+1 dimensional manifold (NOT sure if I used that word correctly) that still maintains the euclidean properties. – Rellek Apr 24 '15 at 20:58
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    You write in your comment "there would always be some point right there before the boundary (ie dotted line) that would always have a neighborhood with points not belonging to the set," but that is false already for the real number line. Do you think there is an $x > 0$ for which you can't conceive of an interval around it that stays entirely to the right of $0$, but rather every possible interval around $x$ must contain negative numbers? (My point here is that $(0,\infty)$ is an open set in $\mathbf R$ and I'm pointing out how that's consistent with the general notion of an open set.) – KCd Apr 24 '15 at 21:16
  • @TravisJ What about the set of all rational numbers $\Bbb Q$ in the discrete topology of $\Bbb R$? When I study general topology, I cannot think of anything in general to make me understand open set other than definition. – MonkeyKing Apr 24 '15 at 21:32
  • @MonkeyKing, if you use the discrete topology (correct me if I'm wrong) the distances between things are always either 0 or 1 (or integral at least). In which case, the small amount you can move in any direction is $0<\delta<1$. – TravisJ Apr 24 '15 at 22:26
  • @TravisJ Let me see. You metricize this space since it is Haussdorf. For non-metricizible space, "wiggle room" idea will makes sense, is it the idea? What a minute, "wiggle room" seems bizarre on topologies of ${a,b,c}$, and I don't think we can metricize most of topologies of this space. Could you help please? – MonkeyKing Apr 24 '15 at 22:39

3 Answers3

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A set is open if, from any point in the set, you can wiggle in any direction a little bit and stay inside the set. What "wiggle" means depends on the context at hand.

In metric spaces, "wiggle" means what you might expect: "move a small distance". That is, for each point in an open subset of a metric space, there is a ball around the point which is contained in the open set.

In one important extreme, the trivial topology $\{ \emptyset,X \}$, there is no wiggle room anywhere: everything is somehow collapsed together, and by wiggling at all you "bump into everything". Formally, any sequence in the trivial topological space converges to every point in the space.

In the other important extreme, the discrete topology, all sets are open, including singletons. In view of how the subspace topology works, a nice way of viewing this is by thinking of a discrete space as a set of isolated points in a larger space that is not itself discrete. For instance $\mathbb{Z}$ is a discrete subset of $\mathbb{R}$. (I am not sure how to give an intuitive explanation of the subspace topology, however.)

Other topological spaces are in between these two extremes.

The axioms of topology make some intuitive sense in this framework:

  • The empty set is vacuously open: my intuitive definition says "from any point in the set" and there are no such points.

  • The entire set is also certainly open, provided we assume that "wiggling" does not take you out of the set.

  • If $U$ is a union of open sets and $x \in U$, then there is a particular open set $O$ in the union with $x \in O$. This $O$ gives $x$ some wiggle room, and this wiggle room is also contained in $U$.

  • If $U$ is a finite intersection of open sets and $x \in U$, then we can take the "wiggle room" of $x$ from each of the sets in the intersection and intersect all of them. This will still leave some room to move around $x$. This is easier to interpret in metric spaces, where the wiggle room consists of balls: the finite intersection of balls of a fixed center is just a ball whose radius is the minimum of all the radii. From this perspective, infinite intersections of open sets needn't be open, because this "minimum" (really, infimum) radius might actually be zero, which would leave us with no room to move.

Some (maybe most) general topology books prove that the definition of a topology with open sets is equivalent to the definition of a topology with a neighborhood system. This latter definition is more closely tied to the intuitive picture I've been trying to draw here.

Note that this idea is not really universal. For example, I don't think there is a way to apply this idea to the Sierpinski space $(\{ 0,1 \},\{\emptyset,\{ 0,1 \},\{ 1 \} \}$).

Ian
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    Thanks for helping me precisify "wiggle room." The only exposure to topology I've had is through real analysis and functional analysis, so it is unfamiliar territory when you can't put a metric on something. The "wiggle room" idea is the way I've always thought about it. Glad to see it can still have some meaning in a non-metrisable setting. – TravisJ Apr 25 '15 at 01:16
  • @TravisJ The weak topology of a topological vector space is typically not metrizable, but this intuition definitely still applies to it. – Ian Apr 25 '15 at 02:00
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I (personally) think of closed sets the way I explained here: How would I explain an 'open set'

One way to think of closed sets is they are the complement of an open set. (In many instances this may even be the definition of a closed set.)

The way I think of closed sets is in terms of convergence of sequences... if you take any sequence $s_{n}\to s$ where each $s_{i}\in S$ then if $S$ is closed you are guaranteed that $s\in S$ too. The "intuitive" way to think of this is that if you are moving around inside a set but slowing down and coming to a stop (eventually a stop... after $\infty$-ly many steps). Once you've slowed down enough, you can predict where you will stop. If it must be the case that your prediction is always a stopping point inside the set, then it is closed. Of course in practice, a sequence could be very bouncy and not come to a stop evenly at all... but this is how I think about it.

Community
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TravisJ
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Topology is literally the study of open sets. A topology is a collection of open sets.

In $\mathbb{R}$, open sets are arbitrary unions of open intervals, $(a,b)$ and closed sets are arbitrary intersections of closed intervals, $[a,b]$. These are important because they define limits, continuity, etc.

In topology, you simply define what "open" means for your space. In that, you define limits, continuity, etc.