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

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?

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 $\mathbb{R}$.