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...*