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I am trying to prove that $\mathbb N$ is closed using the limit point argument ("a set is closed if all it's limit points belong to the set"). In order to achieve the proof by induction, I have to proof this intermediary result :

If two sets have no Limit Points, then their union itself has no LP

In order to prove this, we have to prove that $$\forall p\in \Omega , \exists r>0, \forall x \in S_1\cup S_2, x \notin\mathfrak B_r(p) \cap(S_1\cup S_2)$$

where $S_1,S_2$ are the two sets having no Limit Points, $\Omega$ is the metric space we're in.

So I chose $p\in \Omega$ and $r= min(min_{y\in S_1}d(y,p), min_{z\in S_2} d(z,p))$

I have gotten convinced that with this $r$ it should work, but from there I don't know how to conclude. I feel very uncomfortable with the $\mathfrak B_r(p) \cap(S_1\cup S_2)$..

  1. Is my approach good, or are there any flaws?
  2. If it is good, could you help me to conclude?
niobium
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2 Answers2

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Hint: Use the Archimedian property to construct, for a supposed limit point, a neighbourhood that doesn't intersect the natural numbers.

Edit: As for the proof of the result, apply distributivity to the intersection of the open ball and $S_{1} \cup S_{2}$. However I fail see how that result might help you.

LaxHat1376
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A set is closed if all it's limit points are in it.

Since $N$ has no limit points in the usual topology of $R$ it is vacuously correct that it is closed. Logically:

$(\forall x)(x $ is a limit point of $N\implies x\in N)$.

Since the antecedent is false for all $x$ the condition is true.

ryaron
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  • Yes but how do you prove that $\mathbb N$ has no limit points? The only thing I know is that a finite set has no LP. That's why I decided to go for an induction. – niobium Jan 30 '23 at 11:49
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    @niobium Why would you use induction? . If you want to argue that since finite sets have no limit points then their union cannot have limit points then it would be false (Think of $\Bbb{Q}$). Also what is your definition of a limit point? – Mr.Gandalf Sauron Jan 30 '23 at 11:55
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    A point $p$ is a limit point of $E$ if for every neighbourhood of $p$ there is a point of $E$ different from $p$ in this neighbourhood. (neighbourhoods are open in the text I am studying in) – niobium Jan 30 '23 at 11:58
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    So to prove directly that $\Bbb{N}$ has no limit point, you pick a natural number $n$ and then find an open neighbourhood of it (for example $(n-\frac{1}{2},n+\frac{1}{2})$ which does not contain any point of $\Bbb{N}$ but $n$. And then for any real number $r$, which is not a natural number , you have $\lfloor r\rfloor<r<\lfloor r\rfloor+1$ . Then find an appropriate neighbourhood accordingly. – Mr.Gandalf Sauron Jan 30 '23 at 11:58
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    For example $(r-\epsilon,r+\epsilon)$ where $\epsilon<\min(\lfloor r\rfloor+1 - r , r-\lfloor r\rfloor)$ would work. – Mr.Gandalf Sauron Jan 30 '23 at 12:03
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    @Mr.GandalfSauron It took me like $5$ minutes to understand, but now I got it! Help much appreciate! – niobium Jan 30 '23 at 12:10