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I am trying to prove the above proposition. I tried to prove it by way of contradiction letting $S$ be such nonempty proper subset of $R$. Then $T=R-S$ would also be a nonempty proper subset of $R$ which is both open and closed. The method I thought of to lead contradiction is to create a sequence like this.

Take $s_0$ from $S$ and $t_0$ from $T$. Then consider the point $\frac {s_0+t_0}{2}$ and if this point belongs to $S$ call it $s_1$ otherwise $t_1$. Repeat this process then eventually there is a sequence $s_m$ and $t_m$ which both converge to the same limit. But then since both sets are closed the limit point belongs to both sets, which is a contradiction.

It's clear that such a sequence exists intuitively but I'm having trouble rigorously constructing such a sequence. Since I have no idea which set each of the newly constructed point belongs to, I don't know how to prove by $\epsilon-N$ definition of a sequence that both sequences converge to the same number. Can anybody help me?

Moreover, how may I be able to generalize this proposition to the $R^n$ setting? Thanks.

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marked as duplicate by Andrés E. Caicedo, Najib Idrissi, user127096, Dan Rust, Namaste Apr 21 '14 at 15:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You've got the idea right - just a little bookkeeping needed : Choose $s_0 \in S$ and $t_0 \in T$, and define $r = (s_0 + t_0)/2$ $$ s_1 = \begin{cases} s_0 &: r \in T \\ r &: r\in S \end{cases} \quad\text{ and }\quad t_1 = \begin{cases} r &: r\in T \\ t_0 &: r\in S \end{cases} $$ Now $|s_1 - t_1| = \frac{|s_0-t_0|}{2}$ Now inductively construct the sequence so that $$ |s_n - t_n| \leq \frac{|s_0-t_0|}{2^n} $$ Now check that your argument applies (ie. check that $(s_n)$ and $(t_n)$ both converge).

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  • $\begingroup$ This shows that both sequence must converge to the same limit. But now I'm having trouble showing that both sequences indeed converge to some number. How can I show this part? $\endgroup$ – takecare Apr 21 '14 at 14:48
  • $\begingroup$ You can prove that the sequences are cauchy - hence they must (individually) converge. $\endgroup$ – Prahlad Vaidyanathan Apr 21 '14 at 14:56
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Your idea is fine but may require some technical case distinctions. Alternatively you can (assuming $s_0\in S$, $t_0\in T$, and wlog. $s_0<t_0$) let $a=\inf([s_0,t_0]\cap T)$. As we take the infimum of a nonempty bounded set, $a$ is a real number and in fact $s_0\le a\le t_0$. Investigate the consequences of $a\in S$ or $a\in T$.

A space $X$ with this property to show (i.e. that the only open closed sets are the empty set and the whole space $X$) is called connected. To extend the result to $\mathbb R^n$ one could show that $X\times Y$ is connected if $X$, $Y$ are both connected.

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