Connectedness of a linear continuum

Question

From Munkres pg 153:

Theorem 24.1. If $L$ is a linear continuum in the order topology, then $L$ is connected, and so are intervals and rays in $L$.

Proof. Recall that a subspace $Y$ of $L$ is said to be convex if for every pair of points $a,b$ of $Y$ with $a < b$, the entire interval $[a,b]$ of points of $L$ lies in $Y$. We prove that if $Y$ is a convex subspace of $L$, then $Y$ is connected.

So suppose that $Y$ is the union of the disjoint nonempty sets $A$ and $B$, each of which is open in $Y$. Choose $a \in A$ and $b \in B$; suppose for convenience that $a < b$. The interval $[a,b]$ of points of $L$ is contained in $Y$. Hence $[a,b]$ is the union of the disjoint sets

$$ A_0 = A \cap [a,b]\ \ \textrm{and}\ \ B_0 = B \cap [a,b]\textrm{,}$$

each of which is open in $[a,b]$ in the subspace topology, which is the same as the order topology. The sets $A_0$ and $B_0$ are nonempty because $a \in A_0$ and $b \in B_0$. Thus, $A_0$ and $B_0$ constitute a separation of $[a,b]$.

Let $c = \sup A_0$. We show that $c$ belongs neither to $A_0$ nor $B_0$, which contradicts the fact that $[a,b]$ is the union of $A_0$ and $B_0$.

I do not understand the last sentence, "which contradicts the fact that $[a,b]$ is the union of $A_0$ and $B_0$.

Why does the fact that $c$ belongs neither to $A_0$ nor to $B_0$ give a contradiction?

Answer 1

We know that $Y = A \cup B$ (which is a disjoint union by assumption), so as $[a,b] \subseteq Y$ we know by simple set theory that $$[a,b] = [a,b] \cap Y = [a,b] \cap (A \cup B) = ([a,b] \cap A) \cup (a,b] \cap B) = A_0 \cup B_0\text{.}$$

Then $c = \sup(A_0)$, so $a \in A$ implies $a \le c$. And clearly, as $A_0 \subseteq [a,b]$, $c \le b$ as well ($b$ is surely an upper bound of $A_0$ and $c$ is the smallest one). So $c \in [a,b]$ so $c \in A_0$ or $c \in B_0$ by the above.