# Proof of that every interval is connected

I don't understand the following proof of that every interval is connected in $\mathbb{R}$.

Let $Y$ be an interval in $\mathbb{R}$ and suppose that $Y$ is not connected.

Then $Y=A\cup B$, where $A,B\subseteq Y$ are open in $Y$, $A,B\neq\emptyset$ and $A\cap B=\emptyset$. Let $a\in A$ and $b\in B$. Without loss of generality, $a<b$.

Let $\alpha=\sup\{x\in\mathbb{R}:[a,x)\cap Y\subseteq A\}$.

Then $\alpha\le b$ and $\alpha\in Y$. It is clear that $\alpha\in Cl_{Y}(A)$, and $A$ is closed in $Y$, then $\alpha\in A$. Since $A$ is open in $Y$, $Y$ is an interval, $b\in Y\setminus A$ and $\alpha <b$, then there exists $r>0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$. We can conclude that $[a,\alpha+r)\cap Y\subseteq A$, which is a contradiction.

I don't understand why $\alpha\le b$ and $\alpha\in Y$. Can anyone explain this to me?

Thanks.

If $\alpha$ were strictly greater than $b$, then for all small $\varepsilon>0$ $[a,\alpha-\varepsilon) \cap Y$ would contain $b$. But $[a,\alpha-\varepsilon)\cap Y$ is a subset of $A$, so this would imply $b\in A$. Contradiction, since $A\cap B$ is empty.
$\alpha$ is not necessarily in $Y$, just in its closure. To see this, reduce to cases. If $\alpha\in Y$, we are done. Otherwise, since $\alpha = \sup\{x:[a,x)\cap Y\subset A\}$, there exists a sequence of points $x_n \in\{x:[a,x)\cap Y\subset A\}$ converging to $\alpha$. Since the $x_n$ lie in $Y$, this says there exists a sequence in $Y$ converging to $\alpha$, so $\alpha$ is a limit point and hence in the closure.
• Why the $x_n\in Y$? @neuguy – Solyx91 Sep 17 '14 at 13:10
• In fact, $\alpha$ is in $Y$, because $a\leq \alpha \leq b$, $a,b\in Y$ and $Y$ is an interval. Here is used the property that "Let $Y$ be an interval and $a,b\in Y$. For all $x$, if $a\leq x\leq b$ then $x\in Y$. Note that it is key to the proof that $\alpha$ is in $Y$, otherwise the proof would not be able to proceed to the conclusion that $\alpha\in CL_Y( A)$. – Ramiro Dec 14 '15 at 23:01