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).
 A: 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.
A: Three observations:


*

*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.

*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.

*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
A: 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 $<X,Y>$ 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 $<X,Y>$ the condition $X \neq 0 \neq Y$ implies $X \cap Y \neq 0$. 
A: 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.
