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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.

MyMathYourMath
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Solyx91
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2 Answers2

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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.

Gyu Eun Lee
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    Why the $x_n\in Y$? @neuguy – Solyx91 Sep 17 '14 at 13:10
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    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
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I know this question is very old, but I wanted to give a clearer answer. The current one isn't necessarily wrong, but I think it mixes up the definition of the supremum with the equation defining the set, and requires showing that the set is an interval. Ramiro's comment explains why $\alpha \in Y$ (follows from $Y$ being an interval). Note that $[a, a)$ is empty, so $a \in \{x \in \mathbb{R}: [a, x) \cap Y \subseteq A\}$ and the set is nonempty.

If $z > b$ for some $z \in \mathbb{R}$, then $b \in [a, z)$. Since $b \in B \subset Y$, $b \in Y$ as well, so $b \in [a, z) \cap Y$. Since $b \notin A$, it follows that $z \notin \{x \in \mathbb{R}: [a, x) \cap Y \subseteq A\}$. Hence, $b$ is an upper bound, which means that $\alpha \leq b$ by definition of the supremum.

Vic F.
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