$A'=\mathbb{N}$ for $A=\{{n + 1/m}\mid n,m\in\mathbb N\}$

Asked May 02 '21 at 23:28

Active May 03 '21 at 00:00

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I have to prove in $\mathbb{R}$ that $A'= \mathbb{N} $, where $$A=\{n + 1/m\mid m,n\in\mathbb N\}.$$ Let me first describe my thought on this one.So in order for these 2 sets to be equal it means that $A'=\{1,2,3...\}$ so for $x=1,2,3..$ I have to prove that the ball $B(x,\epsilon)$ with $\epsilon = 1$ will have infinite points of $A$ in it. So we know from Archimedes that we can find $n < 1/n_o$ for every $n,n_o$ real numbers. So i can find infinite points around that $x$.

Is this correct or am I missing something

edited May 03 '21 at 00:00

Thomas Andrews

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asked May 02 '21 at 23:28

sorakisnertis

Do you mean $$A=\left{n+1/m\mid m,n\in\mathbb N\right}?$$ – Thomas Andrews May 02 '21 at 23:38
@ThomasAndrews yes m,n are real numbers – sorakisnertis May 02 '21 at 23:38
Is $A’$ the boundary of $A?$ – Thomas Andrews May 03 '21 at 00:00
@ThomasAndrews in my book A' are all the points lets say $x$ that B(x,e)-{x} intersection A is different from the empty set – sorakisnertis May 03 '21 at 00:05

$A'=\mathbb{N}$ for $A=\{{n + 1/m}\mid n,m\in\mathbb N\}$

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