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In my textbook there is a theorem saying that "If $S\subset\mathbb{R}$ is an interval, then $S$ is connected." I can follow most of the arguments provided there except the one indicated below. Can anyone help me understand it, please? To be complete, I briefly state the proof in my textbook.

$\textit{Proof.}$ Suppose $S$ is an interval in $\mathbb{R}$ and $S$ is not connected. Then there exist nonempty open sets $U$ and $V$ in $S$ such that $$S=U\cup V, U\cap V=\emptyset.$$ Choose $a\in U$ and $b\in V$. Without loss of generality we may assume that $a<b$. Since $S$ is an interval, $[a, b]\subset S$.

$\textit{My question is with the following argument.}$ (Arguments thereafter are omitted.)

Let $c=\mathrm{sup}([a, b]\cap U)$. Since $c\in [a, b]\subset S$, so either $c\in U$ or $c\in V$. Why is $c$ must be in $[a, b]$, please?

LaTeXFan
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1 Answers1

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The set $[a,b]$ is closed, so it contains the supremum and the infimum of each of its non-empty subsets. For a little more detail, let $u=\sup\big([a,b]\cap U\big)$; then by the definition of supremum for each $\epsilon>0$ there is a $u_\epsilon\in[a,b]\cap U$ such that $u-\epsilon<u_\epsilon\le u$. Thus, every open nbhd of $u$ contains a point of $[a,b]$, and $u$ must therefore be in the closure of $[a,b]$. But $[a,b]$ is closed, so $u\in[a,b]$.

Brian M. Scott
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  • Thank you for your help. Just wondering whether the "$\leq$" in the third line should be strictly less. Then every open ball centered at $u$ contains at least a point in $[a, b]\cap U$, i.e. a point in $[a, b]$, which is not $u$. Then $u$ is a limit point of $[a, b]$. – LaTeXFan Nov 16 '13 at 02:52