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...
(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 $1$.
(1) being true immediately implies that (2) and (3) are false.
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 $\mathbb{R}$.
I suggest you read a few proofs in an elementary analysis book (e.g. Introduction to real analysis by Bartle). It will give you an idea of what is required in a proof. In the question, you have only provided numerical examples, which are by no means proofs.