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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.

Mathaddict
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1 Answers1

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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 $m\in \Bbb Z$.

Angina Seng
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