# Are my arguments correct about limit points of $A=${$m+n\sqrt 2:m,n\in \mathbb Z$}? [duplicate]

This question already has an answer here:

Let $A=${$m+n\sqrt 2:m,n\in \mathbb Z$},then-

$(1)A$ is dense in $\mathbb R$.

$(2)A$ has only countable many limit points in $\mathbb R$.

$(3)A$ has no limit points in $\mathbb R$.

$(4)$only irrational numbers can be the limit points of $A$.

$A=${$..,...,-3-3\sqrt 2-2-2\sqrt 2,-1-\sqrt 2,0,1+\sqrt 2,2+2\sqrt 2,3+3\sqrt 2,4+4\sqrt 2,...,...$}

Argument for (1) taking $x=\frac{1}{2}(3+3\sqrt 2)\in\mathbb R-\mathbb A$ taking $\delta=\frac{1}{4}(3+3\sqrt 2)$,then $(x-\delta,x+\delta)\cap A =\phi$.Hence $A$ is not dense in $\mathbb R$

Argument for (2).Since $A$ is not dense in $\mathbb R,$i.e $\bar A\neq \mathbb R,$it means $\bar A$ is either $\mathbb Q$ or $\mathbb Q^c$ or $\emptyset$(please check this point!!!) .Now let us take $q\in \mathbb Q$ and taking $\delta =\frac{1}{4}(m+(n-1)\sqrt 2)$,then $(q-\delta,q+\delta)\cap A=\phi$.Hence $\bar A\neq \mathbb Q$.Hence,$A$ does not have countable many limit points in $\mathbb R$

Argument for (4) taking $x=\frac{1}{2}(3+3\sqrt 2)\in\mathbb Q^c$ taking $\delta=\frac{1}{4}(3+3\sqrt 2)$,then $(x-\delta,x+\delta)\cap A =\emptyset$.Hence,only irrational numbers cannot be the limit points of $A$

Hence the only possibiliy left is $\emptyset$.So, $A$ has no limit points,making option (3) true.

Please check my arguments,whether they are correct or not?

Also,please suggest if some improvements can be made in above arguments

My question is not duplicate of any question.In the suggested duplicate,the proof is given,but i don't want the proof of this,i just want to clarify my concept via this problem regarding limit points,i just wanted to check my arguments,whether they are correct or not...

## marked as duplicate by carmichael561, Misha Lavrov, Robert Soupe, The Phenotype, Parcly TaxelFeb 23 '18 at 9:25

• For (1), how do you obtain $(x-\delta,x+\delta)\cap A =\emptyset$?. What about the point $5-2\sqrt{2} \in A\cap (x-\delta,x+\delta)$ – Quoka Feb 23 '18 at 6:56
• Please use \emptyset or \varnothing for the empty set. It looks a lot better, and specifically doesn't look like a Greek letter that might represent some other set. Also, how do you figure that $(x-\delta,x+\delta)\cap A =\varnothing$? How can you tell so easily that something like $1411-1000\sqrt2$ isn't in that interval? – Arthur Feb 23 '18 at 6:58
• @MathUser_NotPrime:Sorry,it was a typo.Now see the edit. – P.Styles Feb 23 '18 at 7:01
• It looks like you only considered the case when both $m,n$ are positive or both negative. – N. S. Feb 23 '18 at 7:06
• @N.S.:you marked my misunderstanding,thank you – P.Styles Feb 23 '18 at 7:07

(1) is true while (2), (3), (4) are false. To see that (4) is false, note that since $1\in A$ the constant sequence $1$ is a sequence in $A$ converging to the rational number $1$.
It remains to prove (1). This question has been asked multiple times. A solution is provide here : Proving that $m+n\sqrt{2}$ is dense in R.
Now, you wrote: Since $A$ is not dense in $\mathbb{R}$, it means that $A=\mathbb{Q}$ or $\mathbb{Q}^\complement$. This is certainly not the case. Consider for instance the interval $(0,1)$, or the set $\{1\}$. Neither of these sets are dense in $\mathbb{R}$ but they are neither $\mathbb{Q}$ nor $\mathbb{Q}^\complement$. Furthermore, both $\mathbb{Q}$ and $\mathbb{Q}^\complement$ are actually dense in $\mathbb{R}$.