# Is there such thing as a continuous set?

I've been working with sets recently and have often thought about the idea of set continuity. However, I don't remember ever being taught about set continuity or anyone else mentioning it (I've even received push-back). After investigating for a little, I the only reference to "continuous set" I've found is from the Encyclopedia of Mathematics. Which has a definition for a continuous set and then says: "The phrase "continuous set" is not used in the Western literature".

So my question is: is there such a thing as a continuous set? If so, where can I find a definition for it (other than EoM)? If not, then why is the study of this property unimportant/inconsequential?

## Here is my proposed definition of what a continuous should be:

Set $$A$$ is said to be continuous $$\Leftrightarrow$$ $$\forall B\neq\emptyset, B\subset A, \exists \ C \ such \ that\ C \cap B\neq\emptyset, B \nsupseteq C\subset A$$ where $$C$$ is a convex set not contained in $$B$$.

Put simply, a set is continuous if it is possible to travel from any point in the set to any other point in the set without leaving the set.

One Particular example:

I've been studying preference continuity in microeconomics, and it seems obvious to me that preferences must be defined in a continuous set in order to be continuous themselves. But I think this is often left to one side as preferences are commonly thought in $$\mathbb{R}_+^L$$, which is a continuous set (However I am more concerned on the mathematical use of a continuous set than it's use in this particular example).

• Are you aware of the notion of connectedness, and in particular path connectedness?
– kccu
Sep 20 '19 at 0:33
• In addition to @kccu's comment on your informal characterization, note that your formal characterization doesn't seem to work: the only "continuous" set would be the empty set (and any obvious change would result in "continuous" simply being "convex") Sep 20 '19 at 0:43
• If you can write down a proof that $\mathbb{R}_+^L$ is "continuous" in the sense that you're thinking of, that will help to clarify what you really want the definition to be. For that matter, even a simple example like $\mathbb{R}$ would help. Sep 20 '19 at 0:49
• This is a good question. I don't know why there is a vote to close. Sep 20 '19 at 1:08
• @cachdies Also, you say you think a complete metric space should be "continuous"? What about the metric space $X=[0,1] \cup [2,3]$? What about $\mathbb{Z}$? These are complete metric spaces, but they don't seem to have the properties you want.
– kccu
Sep 21 '19 at 13:07

Three observations:

1. It seems like you mean to require $$B$$ to be a proper subset of $$A$$, but that condition isn't stated. As Brian Moehring points out, when $$B=A$$ and $$A$$ is nonempty, there does not exist any such $$C$$, which trivializes the definition.
2. From the "put simply" comment, it seems like you mean to talk about partitioning $$A$$ into two subsets. If so, it's better to respect that symmetry, and to phrase your definition in those terms, not in terms of $$B$$ and its complement.
3. Ease up on the symbols! They're hard to read.

With that said, here's an improved definition:

• Let $$A$$ be a subset of a real affine space. We say that $$A$$ is continuous if for every partition $$A=X\sqcup Y$$ into two nonempty subsets, there exists a line segment in $$A$$ that intersects both $$X$$ and $$Y$$.

Is that what you meant?

Finally, about the choice of the word itself. As you know, "continuous" is overloaded, so let's avoid that. The definition isn't the same as "convex", nor is it obviously any kind of "[noun]-convex". And the definition turns out to be about bridging cuts, not connecting pairs of points, so I wouldn't call it "[noun]-connected" either. How about something new: stitchable?

Edit: I take back what I said about ruling out "[noun]-connected". I believe your definition may be equivalent to polyline-connected! The latter phrase is uncommon, but its meaning is straightforward enough, and it is treated briefly in the AMS Elementary Topology: Problem Textbook , see https://books.google.com/books?id=7U8-rs-S2boC&pg=PA95

• Thanks Chris! 1. You are right, I have edited the original formula so $B$ is a popper subset of $A$. 2. I'm not sure I understand entirely, I guess that it is possible to phrase it in some way as to refer to $B$ and it's complement but I fail to understand it's benefits. 3. ó_ò - sorry, but I hope it's understandable. I like your definition, I believe we are referring to the same thing. I like the idea of calling it "stitchable" but I believe it is the same as Connectedness. polyline-connected is a very useful notion thanks for the reference and the link! Sep 20 '19 at 20:22
• 2. Your conditions on $C$ amount to saying that $C$ must intersect both $B$ and its complement. But the second part is phrased in a roundabout way, saying instead that $C$ should not be a subset of $B$. Have some pity for your human readers, for whom it is not at all clear why anyone should care whether $C$ is a subset of $B$! Note that I am not suggesting that you should explicitly refer to the complement of $B$; I'm saying you should go one step further and give it a name. That's why I name both $X$ and $Y$. Sep 20 '19 at 20:44
• 3. This is not the same thing as topological connectedness. The circle, as a subset of $\mathbb R^2$, is connected. But it is not polyline-connected; in fact, no nontrivial subset of the circle is convex. Sep 20 '19 at 20:45
• You are right, if you refer to "circle" rather than "disk". I guess "stitchable" does require the set to be polyline-connected🤔. Sep 20 '19 at 21:39

Special thanks to @chris-culter and @kccu. The word "continuous" is not commonly used to describe sets, instead sets are said to be a Connected space.

The proposed definitions by myself and Chris Culter instead require a set to be polyline-connected, which is in it's self is a relatively obscure concept. This is because connected spaces don't necessarily contain a convex set (equivalent to $$C$$), think a circle vs. a disk.

Moreover, the reason sets are said to be connected rather than continuous is, as kccu noted, because "Connectedness is a property of sets. Continuity is a property of functions. Therefore the notion of connectedness is used for sets, while the notion of continuity is not. You could call connectedness "continuity" if you wanted to, but that would probably be confusing since you're overloading the term"

I hope that if anyone else has the same question as I did, they'll be able to find this post and come to the same understanding I have.

• However, I'm proud of my definition and believe if the restriction of $C$ to be convex, where to be relaxed to be connected, then connectedness and continuity or "stitchable" would be the same. Sep 20 '19 at 22:00

The following definition of the continuos set is given by K. Kuratowski in his classic "Set theory" (Chapter 6).

Consider a linearly ordered set $$A$$ and its cuts, i.e., such pairs $$$$ of $$A$$'s subsets that $$X=Y^-$$ and $$Y=X^+$$, where $$U^{-(+)}$$ denotes the set of all elements $$\leq (\geq)$$ than each element from $$U$$.

Example of a cut. $$<\{a\}^-,\{a\}^+>$$ (for all $$a \in A$$)

$$A$$ is called continuous, if for all cuts $$$$ the condition $$X \neq 0 \neq Y$$ implies $$X \cap Y \neq 0$$.

You've tagged your post elementary-set-theory. But “continuous” is a term that is used in connection with topology, which is a bit more structure than merely set theory.

Basically, in set theory, the set elements are treated as completely independent from each other, while in topology, there is also a certain notion of ”closeness” available.

In your proposed definition, you go even further and use a convex set. That requires even more structure than topology. The usual definition actually requires a linear space.

Your “put simply” statement actually describes a topological property known as path-connectedness. Note that path-connectedness does not need convex sets; for example, a circle is path-connected, but the only non-empty convex sets in a circle are singletons (sets with exactly one element).

By definition, a set is path-connected if for any two points $$p,q$$ it contains a path connecting those points. Such a path, in turn, is a continuous function $$f$$ from the interval $$[0,1]$$ to the space in question with $$f(0)=p$$ and $$f(1)=q$$.

There is also the related, but slightly weaker concept of a connected set, which in essence says you cannot split the space into two parts that don't share a border.

Note that the concept of connectedness is purely topological, so you don't need any structure beyond topology. But you still need topology.