How would I prove that $\mathbb{N}$ has no limit points? Why does $\mathbb{N}$ have no limit points?

I have tried proving with integers but clearly this is not the same.


closed as off-topic by Namaste, Saad, Xander Henderson, Shailesh, Mohammad Riazi-Kermani Mar 28 '18 at 6:29

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    $\begingroup$ Is $N$ the natural numbers? If so, the argument is very similar as the argument for the integers. $\endgroup$ – FYY Mar 6 '18 at 2:17

Let us suppose $\mathbb{N}$ has a limit point say $a$. Then, for any $\varepsilon$>0 $\exists$ an open neighborhood $\eta=(a-\varepsilon,a+\varepsilon)$ s.t. $\eta-\{a\}\cap\mathbb{N}\neq\emptyset$. Which is a contradiction as $\mathbb{N}$ contains no points other than integers. So $\mathbb{N}$ has no limit points.


For a point $x$ to be a limit point, we need a sequence {$x_n$} where $x_n \ne x$ in our set to approach $x$.

That does not happen for the set of natural numbers because points are at least one unit apart from each other.


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