Baby Rudin gives the example of the set of all integers being not open if it is a subset of $\mathbb{R}^2$.

If we consider the set of integers in $\mathbb{R}$, is this set also not open? I can find a neighbourhood which will contain any point, $p$, however is it a requirement that a neighbourhood contains more than one point?

I'm trying to understand this fully and have searched through the various posts that have a slight relation and can not find out specifically how these take interior and isolated points into account and how these relate to openess.

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    $\begingroup$ The set of integers is not an open set in $\mathbb R$. In the real numbers, no neighborhood contains a single point. $\endgroup$ – Thomas Andrews Feb 13 '16 at 14:06
  • $\begingroup$ You are correct. $\mathbb{Z}$ is not an open subset of $\mathbb{R}$. But your argument in the second paragraph doesn't really say why that's true. I think you know what you are trying to say, but haven't said it. $\endgroup$ – Ethan Bolker Feb 13 '16 at 14:07
  • $\begingroup$ $\mathbb{Z}$ is not open in $\mathbb{R}$. And if in addition you were to ask if $\mathbb{Z}$ is closed or not closed in $\mathbb{R}$ the answer would be that $\mathbb{Z}$ is closed in $\mathbb{R}$. $\endgroup$ – Jeppe Stig Nielsen Feb 13 '16 at 16:42
  • $\begingroup$ @Jeppe Stig Nielson Yeah, I understand the existence of limit points and how that relates. My only issue was openess and isolated points really. There was the other question of the requirement that a neighborhood have more than one point for it to be a neighborhood. Which I assume is a requirement. Therefore you can't have a neighborhood of a singleton set. $\endgroup$ – smokeypeat Feb 13 '16 at 16:58

A set $U\subset \mathbb R$ is open if and only if for every $x\in U$, there exists some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is a subset of $U$.

For $U=\mathbb Z$, this is clearly not the case:

  • Take $x=0$
  • Take any $\epsilon > 0$.
  • Then, $\min\{x+\frac\epsilon2, x+\frac12\}$ is an element of $(x-\epsilon, x+\epsilon)$, but it is not an element of $\mathbb Z$.
  • Therefore, $(x-\epsilon, x+\epsilon)$ is not a subset of $\mathbb Z$ for any value of $\epsilon$
  • Therefore, $\mathbb Z$ is not open.
  • $\begingroup$ Would you then call each integer in the set isolated and then would it be wrong to generalize a set of isolated points as being not open? $\endgroup$ – smokeypeat Feb 13 '16 at 15:30
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    $\begingroup$ @inandouto'mind It's even worse: a set with any isolated point is not open. For example, $(0,1)\cup \{2\}$ is not open. $\endgroup$ – 5xum Feb 13 '16 at 16:01

$\mathbb{Z}$ is not open in $\mathbb{R}$.

One way to see this is that given any $n\in \mathbb{Z}$ we have for every $\epsilon>0$ that $(n-\epsilon, n+\epsilon)$ is not contained in $\mathbb{Z}$.


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