As the topic, prove that Intervals are connected and only connected in $\mathbb{R}$. I know what is the definition of connected set. But not sure how to prove that.
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2Maybe some of the discussion here is helpful? Set $A$ interval in $\mathbb{R}\implies$ connected – yunone Nov 17 '12 at 05:56
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1In Principles of Mathematical Analysis by Rudin, a proof for the same is given. – Gautam Shenoy Nov 17 '12 at 05:58
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$\newcommand{\cl}{\operatorname{cl}}$HINTS: Suppose that $A\subseteq\Bbb R$ is not an interval; then there are points $a,b\in A$ and $x\in\Bbb R\setminus A$ such that $a<x<b$. Use the sets $A\cap(\leftarrow,x)$ and $A\cap(x,\to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $\Bbb R$ such that $A\cap U\ne\varnothing\ne A\setminus U$ and $A\cap U= A\cap\cl U$; why? Fix $a\in A\cap U$ and $b\in A\setminus U$ and show that $[a,b]\nsubseteq A$, so that $A$ cannot be an interval.
Brian M. Scott
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1Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open – Mathematics Nov 17 '12 at 06:21
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@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $A\cap U$ and $A\setminus U$ are both relatively open subsets of $A$. – Brian M. Scott Nov 17 '12 at 06:23
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$A\cup U\ne\emptyset\ne A\setminus U\implies A\cup U \ne A\setminus U$?I am a little bit confused with the notation – Mathematics Nov 17 '12 at 06:37
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1@Mathematics: No, it doesn’t. But it’s true that $A\cup U\ne A\setminus U$, simply because $U\ne\varnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $\Bbb R$ such that $A\cap U$ and $A\setminus\cl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$. – Brian M. Scott Nov 17 '12 at 06:40
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This proof doesn't seem to cover singleton sets. Am I missing something? Singleton sets are connected sets in $\mathbb{R}$, no? – William Aug 29 '22 at 23:24
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2@William: Yes, singletons in any topological space are connected. And in $\Bbb R$ they are also intervals: ${x}=[x,x]$. The argument is complete, once the details are filled in: the first part shows that every connected subset of $\Bbb R$ is an interval, and the second part shows that every interval in $\Bbb R$ is connected. – Brian M. Scott Aug 30 '22 at 00:57
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For one way, to prove every connected subset of $\mathbb{R}$ is interval, Brian's method works.
For the converse, there is an even simpler way:
Given an interval in $\mathbb{R}$, we can prove that it is path-connected. Given two points in an interval, we can define the path as the line between them.
Path-connectivity implies connectivity.
