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If we consider the open set $\mathbb{R}$, then for every $a \in \mathbb{R}$, you can find an open interval $(a-\epsilon, a+\epsilon)$.

I am probably over thinking this, but I am wondering: Why would it be an open interval if the boundary points are elements in $\mathbb{R}$? I know that $\mathbb{R}$ is both closed and open, but I don't see how the intervals would be open.

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Why would the boundary points being elements of $\mathbb R$ be a problem? Where else could the boundary points be besides $\mathbb R$? –  Jonas Meyer Dec 7 '12 at 3:38
    
I guess what I mean is,shouldn't it be a closed interval instead of an open one? –  Alti Dec 7 '12 at 3:39
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Alti: It is true that $[a-\varepsilon,a+\varepsilon]$ is also a subset of $\mathbb R$, but that takes nothing away from (and it even implies) the fact that $(a-\varepsilon,a+\varepsilon)$ is a subset of $\mathbb R$. –  Jonas Meyer Dec 7 '12 at 3:40
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Actually, it doesn't matter. Let $S$ be a set of reals. Then there exists $\epsilon\gt 0$ such that $(a-\epsilon,a+\epsilon)$ is a subset of $S$ iff there exists $\epsilon'\gt 0$ such that $[a-\epsilon',a+\epsilon']$ is a subset of $S$. I prefer the open interval characterization, but for $\mathbb{R}$ it doesn't matter. –  André Nicolas Dec 7 '12 at 3:43
    
Okay, that helps clarify things, thank you both! –  Alti Dec 7 '12 at 3:45

1 Answer 1

An open interval $(a,b)$ is open because for every $x\in (a,b)$, there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset (a,b)$.

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Thanks,but my main confusion is why is $(x-\epsilon, x+\epsilon)$ open if the boundary points are elements in $(a,b)$? Shouldn't it be a closed interval? –  Alti Dec 7 '12 at 3:42
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While it makes no difference for that interval, note that it matters that $(a,b)$ is open... –  N. S. Dec 7 '12 at 3:47
    
Yes, I was just over analyzing the idea of the open ball being "open" when it could be a closed interval. I got it now, thanks. –  Alti Dec 7 '12 at 3:53

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