Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

3 Answers

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 $\bar B(x^{(0)},r):=\{x\in\Bbb R^d,\max_{1\leq j\leq d}|x_j-x_j^{(0)}|\leq r\}$.

share|improve this answer
    
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
show 1 more comment

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 $a$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.