# Prove $C = \{ x \mid x < a \} \cup \{a\} \cup \underline{ab} \cup \{b \} \cup \{ x \mid b < x \}.$

If $\underline{ab}$ is a region in $C$, then: $C = \{ x \mid x < a \} \cup \{a\} \cup \underline{ab} \cup \{b \} \cup \{ x \mid b < x \}.$

Where C is a continuum that is nonempty, has no first or last point, and is ordered $<$.
Regions can be defined as all the points between $a$ and $b$ (such that $a<b$) denoted by $\underline{ab}$.

This seems a bit obvious to me, but perhaps the proof is more clear. I thought of trying to prove each possible point would end up being some point on $C$ and that $\underline{ab}$ is also a continuum, but I'm not sure this is the way to go.

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Let $X$ be $\{x|x<a\}\cup \dots \cup \{x|b<x\}$.

Clearly $X \subset C$.

Let's take a point $y\in C$. Then either $y<a, y = a, y> a$.

If $y<a$, then $y \in X$.

If $y = a$ then $y \in X$.

If $y > a$, then either $y<b, y = b, b<y$.

If $y <b$ then $a<y<b$ and so $y \in \underline{ab}$, so $y \in X$.

If $y = b$ then $y \in X$.

If $y > b$, then $y \in X$.

So no matter what point you take from C, it is in X. Therefore $C \subset X$, $C=X$.

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I also think the proof should be this clear and simple. – Casquibaldo Sep 30 '12 at 23:09

This is true for any linearly ordered set $(C,<)$, where linearly ordered means that

• $<\$ is transitive
• for all $x,y$ either $x<y$, or $x>y$, or $x=y$.
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Yes, it's true, but to prove it, would one have to do something more elaborate than just proving a few cases? – Casquibaldo Sep 30 '12 at 21:48
I don't think so. – Berci Sep 30 '12 at 23:02
Ok, sounds good. – Casquibaldo Sep 30 '12 at 23:10