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As the topic, how to prove that the only set in $\mathbb{R^1}$ which are both open and close are the $\mathbb{R^1}$ and $\emptyset$. I tried to prove by contradiction, but i can't really show that the assumption implies the contrary.

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Suppose that $X$ is both open and closed and that $a\in X$. Let $S = \lbrace x\ge a: [a,x]\subset X\rbrace$. Is $S$ bounded above? –  Sean Eberhard Sep 23 '12 at 11:24
not sure if S is bounded or not –  abc Sep 23 '12 at 11:30

3 Answers 3

Let $S\subset \Bbb R$ non-empty, open and closed. Fix $x_0\in S$. Let $I:=\{r>0,[x_0-r,x_0+r]\subset S$. As $S$ is open, $I$ is non-empty. If $I$ is bounded, let $\{r_n\}$ a sequence which increases to $\sup I$. Then $x_0+r_n\in S$ for each $n$, and as $S$ is closed $x_0+\sup I\in S$. But $S$ is open, so we can find $\delta>0$ such that $x_0\pm \sup I\pm t \in S$ for $0\leq t\leq \delta$, hence $\sup I+\delta\in I$, a contradiction.

So $S=\Bbb R$.

Note that such an approach works for $\Bbb R^d$ instead of $\Bbb R$. Just replace the interval $[x_0-r,x_0+r]$ by the closed ball $\bar B(x^{(0)},r):=\{x\in\Bbb R^d,\max_{1\leq j\leq d}|x_j-x_j^{(0)}|\leq r\}$.

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If $I$ is unbounded, can you still get a contrary or you just show it is equal to $\mathbb{R^1}$?? –  abc Sep 23 '12 at 11:45
It shows that $I=\Bbb R^1$, as $I$ is an interval. –  Davide Giraudo Sep 23 '12 at 11:48
So, is it true that $S$ is bounded since $I$ is bounded? –  abc Sep 23 '12 at 11:51
Actually, $I$ is not bounded (I get a contradiction when I assumed it, and I think it's what you meant in your comment). –  Davide Giraudo Sep 23 '12 at 11:52
o yup, also, for the part about sequence {$r_n$},, do you mean that $I$ is closed instead of $n$ is closed? –  abc Sep 23 '12 at 11:56

Assume $U$ and its complement $V$ are both non-empty open subsets of $\mathbb R$. Then there are $x \in U$ and $y \in V$ and by switching the roles of $U$ and $V$, if necessary, we may assume $x < y$. Now let $$a = \sup\{b \in \mathbb R : [x,b] \subseteq U \}$$ (the supremum exists since $x \in U$ and $y \not\in U$ and $x<y$). If $a \in U$ then, since $U$ is open, $a + \varepsilon \in U$ for small $\varepsilon$, contradicting the definition of $a$. Otherwise, if $a \in V$ then, since $V$ is also open, $a-\varepsilon \in V$ for small $\varepsilon$, again contradicting the definition of $a$.

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If $U$ is unbounded, $a=+\infty$ so how to get the contrary –  abc Sep 23 '12 at 11:47
Even if $U$ is unbounded, $a$ isn't its $\sup$. $a$ is the $\sup$ of all $b \in \mathbb{R}$ for which the interval $[x, b]$ is a subset of $U$. This set must be bounded as $y \notin U$ and $x < y$. –  Ayman Hourieh Sep 23 '12 at 11:55
o, i c. So if U is unbounded and is a subset of R, would it still implies contrary? –  abc Sep 23 '12 at 12:02
@abc: Boundedness is irrelevant. The attention is focussed on the interval $[x,y]$ only, where there is a switch (at least one) from "belonging to $U$" to "not belonging to $U$". –  Marc van Leeuwen Sep 23 '12 at 14:02

Take a set $A\subseteq \bf R$ which is closed and open. Suppose towards contradiction that $A$ is not the entire $\bf R$ and nonempty. Then there is some point $x_1\notin A$ and $x_0\in A$. Without loss of generality $x_0<x_1$.

Consider the interval $I=[x_0,x_1]$. $I\cap A$ is an intersection of closed sets, so it is closed, so $x'=\sup(I\cap A)$ is in $I\cap A$. Obviously, $x'<x_1$. But $A$ is open, so there is an $\varepsilon>0$ such that $(x'-\varepsilon,x'+\varepsilon )\subseteq A$, but then there's some other point of $A$ in $(x',x_1]$, so we have a contradiction.

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Nice Answer :-) –  Wanderer Sep 30 '14 at 9:38

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