This question already has an answer here:

let $A = \{m + n \sqrt{2}\}$ where $m,n$ are integers, then

$a.$ $A$ is dense in $R$.

$b$. $A$ has no limit point in $R$.

$c$. $A$ has only countably many limit points in $R$.

$d$. only irrational numbers can be limit point of $A$

Set $A$ is similar to the set $N \times N$, so all points are isolated in $A$.
therefore option $b$ should be correct. In answer key, option $a$ is marked as correct. Please suggest if I am doing something wrong.


marked as duplicate by Eevee Trainer, Lord Shark the Unknown, Saad, Paramanand Singh real-analysis Dec 27 '18 at 8:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You say "Set $A$ can be represented as $(m,n\sqrt2),\ldots$". $\endgroup$ – Lord Shark the Unknown Dec 27 '18 at 5:43
  • $\begingroup$ I edited the question! $\endgroup$ – Mathsaddict Dec 27 '18 at 5:46
  • $\begingroup$ You now say "Set $A$ is similar to the set $\Bbb N\times\Bbb N$". Similar? How? $\endgroup$ – Lord Shark the Unknown Dec 27 '18 at 5:47

Consider $\alpha=\sqrt2-1$. Then $\alpha^n\in A$ for all $n\in\Bbb N$ (why?) and $\alpha^n\to0$. So $0$ is a limit point of $A$, which contradicts (d).

In general, if $I$ is an open interval in $\Bbb R$ then $\alpha^n$ is less than the length of $I$ for some $n\in\Bbb N$, and then $m\alpha^n\in I$ for some $m\in \Bbb Z$.

  • $\begingroup$ Thanks, I got it. Can't this set be similar to $Q \times Q$ or $N \times N$ in terms of cardinality? $\endgroup$ – Mathsaddict Dec 27 '18 at 6:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.