A problem related to continuous interval

Suppose $S = (a,b)$ and $T = (c,d)$, where:

1. $a < d$.
2. Both $S$ and $T$ are non-empty.
3. $S \cap T = \emptyset$.
4. $S \cup T$ is connected.

Now the question is: What is $b - c$?

The paradox is that, it can't be $b - c \leq 0$, because then $b$ is not included in $S \cup T$, so it's not connected. But it can neither hold that $b - c > 0$. Because if we then set $\epsilon = b - c$, $c + \epsilon / 2$ is included in both $S$ and $T$, so $S \cap T \neq \emptyset$.

Exculpate myself from disdain: Maybe at first glance everyone regards this as a trivial problem. Yes, it is, somehow. Because I'm not majoring in math and these are just random thoughts. I just want to examine what will happen when you put two open sets on the real axis close enough, and what's the critical point where they merge into a single, bigger one. All these thoughts are based on pure, plain intuition. I don't know too much definitions/axioms/theorems about real analysis. I never had that course.

Perhaps that's why this problem looks stupid. Unfortunately, I finally found the idea I wanted to express is just how connectedness is defined (also pointed out by one of the answers below). So there is no chance to win, as definition cannot be violated. But I think the idea and motivation in it is clear to everyone, and you don't need a definition to think of it. The mystery is still there: You move two disjoint open intervals closer and closer, and finally they intersect. But you don't know where! The definition itself doesn't answer it. Also, you cannot have too many definitions, because the more definitions, the more likely there is hidden inconsistency amongst them. So this problem is solved in terms of math, and I don't think anyone can solve it in terms of logic and philosophy. So I'll mark this problem as answered. Thanks for all the answers. Very helpful, indeed.

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what does "continuous" mean? I expect the interpretation of "continuous" makes 3 inconsistent with 1 and 2 – m_t_ Jul 28 '11 at 15:53
@mt_: it will also depend on whether open intervals have to be nonempty, I suspect. – Carl Mummert Jul 28 '11 at 16:00
There are a few things that are unclear with your question. (a) What does it mean for a set to be "continuous"? Do you mean "connected"? (b) What does it mean for a set to be equal, or not equal, to $\phi$? Perhaps you mean $\emptyset$? The command for $\emptyset$ is \emptyset, not \phi. (c) What do you mean by "obvious inconsistency"? – Willie Wong Jul 28 '11 at 16:07
Voting to close. If you defined your terms properly, the answer would be trivial. – TonyK Jul 28 '11 at 16:39
@Carl is right. You must have either $b \le a$ or $d \le c$ in order for the disjoint union to be connected. – GEdgar Jul 28 '11 at 16:40

Ok, here's how I would broach this topic. Note that both S and T are open sets. I will assume that you know that the union of two open sets is an open set. Now you're requirements state that you have the union of two disjoint open sets is the entire set we're dealing with, and it's also open. This happens to be a definition of connectedness. Actually, it's the definition of everything that is not connected.

Now I don't know what you might mean be continuous, because continuity tends to refer to some sort of function doing some sort of operation. Perhaps a continuous function is one that takes open sets to open sets, maybe. But I will assume that you mean it in the, I can place my pencil on part of my interval and cover all the interval without picking up my pencil, sort of way. But this is the intuitive way of thinking of connectedness.

So what your statements say, then, is that you have a disconnected set that is also connected. No such set exists, so the only set that satisfies these requirements is the empty set. So ultimately, they are inconsistent.

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Just for your information: The French adjective continu (continuous) was used as a noun un continu in the Polish topology/descriptive set theory school - e.g. by Sierpiński and Kuratowski - to mean compact and connected space. – t.b. Jul 28 '11 at 19:55
@Theo: Way cool! Thanks Theo. – mixedmath Jul 28 '11 at 21:06
At the end of my MO question here you'll find two references to back that claim up. By the way, the continuum hypothesis (German Kontinuumshypothese) is of course closely related, the continuum referring to the real line here, of course. (end of today's history lesson, promised). – t.b. Jul 28 '11 at 21:12

The problem is best addressed by drawing a couple of pictures. If there is overlap between $(a,b)$ and $(c,d)$, then your Condition $2$ is violated, since $S\cap T \ne \emptyset$. If there is no overlap, then $S \cup T$ is not connected, and therefore Condition $3$ is violated.

The above is essentially what you wrote. So you proved that Condition $2$ and Condition $3$ are incompatible. I agree that it takes a little thinking to show that the conditions are incompatible. The incompatibility does not arise merely from some sort of formal manipulation involving the symbols $S$, $T$, $\cup$, and $\cap$. What $S$ and $T$ represent is also relevant. For example, if we make a very small change, from $(a,b)$ to $(a,b]$, the incompatibility disappears.

But you did do the thinking required to prove the incompatibility of Conditions $2$ and $3$. There is no paradox.

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