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.
Best Answer
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.