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Please don't be upset with me for asking yet another question about the proof that an interval in $\mathbb{R}$ is complete. From the comments on my last question on the subject I now realise how to prove this in a satisfying way, thank you! But, the proof given by my professor still confuses me, so I wonder about this proof as well. It is really one step I don't understand:

Prop: The interval $(0,1)$ is connected.

Proof: Suppose that $(0, 1) = U\bigcup V$ where $U$ and $V$ are disjoint nonempty sets. We will show that $U$ and $V$ cannot both be open. Let $u \in U$ and $v \in V$ and suppose that $0 < u < v < 1$ (if not we may relabel U and V ). Now consider the set $S = \{x \in U : x < v\}$ This is nonempty since $u \in S$ and bounded above by $v$, so by the Completeness axiom for the real numbers there exists a least upper bound $l = \sup (S)$. We have $u \leq l \leq v$ so $l \in (0, 1) = U \bigcup V .$ Suppose $l \in U$ and let $r > 0.$ If $[l, l + r) \subseteq U$ then $l + \frac{r}{2} \in S$ (WHY?) which contradicts the fact that $l = \sup (S).$ It follows that no open ball $(l - r, l + r)$ is contained in $U$ and hence $U$ is not open in $(0, 1).$

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If $[l,l+r)\subseteq U$, then $l+\frac{r}2\in U$, so it only remains to show that $l+\frac{r}2<v$. Recall that the whole interval $[l,l+r)$ is contained in $U$, that $v\ge l$, and that $v\notin U$; $v$ cannot be in $[l,l+r)$, and $v\not<l$, so it must be true that $v\ge l+r$ and hence that $l+\frac{r}2<v$.

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  • $\begingroup$ you have been faster than me, +1. A final remark, although probably not necessary, $v \notin [l,l+r)$ because $U$ and $V$ are supposed to be disjoint. $\endgroup$ – user67133 May 7 '13 at 17:29

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