3
$\begingroup$

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.

  • 0
    Why would the boundary points being elements of $\mathbb R$ be a problem? Where else could the boundary points be besides $\mathbb R$?2012-12-07
  • 0
    I guess what I mean is,shouldn't it be a closed interval instead of an open one?2012-12-07
  • 1
    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 $\mathbb R$.2012-12-07
  • 1
    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.2012-12-07
  • 0
    Okay, that helps clarify things, thank you both!2012-12-07

1 Answers 1

3

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

  • 0
    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?2012-12-07
  • 1
    While it makes no difference for that interval, note that it matters that $(a,b)$ is open...2012-12-07
  • 0
    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.2012-12-07