1

My solution:

There is no accumulation point in this set. But suppose, 2.4 is accumulation point of this set. But $N_{\epsilon}(2.4)\cap(X without (2.4)) = {} $ So, there is no accumulation point.

Set is not open, because $\epsilon$-Neighbourhood near 1+sqrt(2) (1+sqrt(2)- $\epsilon$,1+sqrt(2)+ $\epsilon$) is not in our set.

But what about closed set? If set don't have accumulation points, what can i conclude about closeness of set?

And chech, if my logic is correct(open set< accumulation point)

Daniil Yefimov
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2 Answers2

5

No, there are "a lot of" accumulation points (I guess that $m,n\in\mathbb{Z}$).

Hint. Note that there is a sequence of rational numbers which tends to $\sqrt{2}$.

Robert Z
  • 145,942
  • $m + n\sqrt 2$ is irrational number. Than we need to prove, that between each pair of irrational number exists rational number. And then, we can use the rational number as our epsilon-neighbourhood to prove, that there are infinitely many accumulation points. Am i right?

    Because I don't understand, why do we need to show, that $(\sqrt2-1)^n\to \infty $

     – Daniil Yefimov Oct 18 '16 at 15:55
  • @Daniel Yefimov $(\sqrt2-1)^n\to 0$ means that the set $G:={ m + m\sqrt{2} \mid m, n \in \mathbb{Z} }$ has $0$ as accumulation point. Now $G$ is a subgroup of $(\mathbb{R},+)$ and the existence of an accumulation points implies that $G$ is dense in $\mathbb{R}$. See here: http://math.stackexchange.com/questions/90177/subgroup-of-mathbbr-either-dense-or-has-a-least-positive-element – Robert Z Oct 18 '16 at 16:13
  • but know i understood the approach. Thank you very much – Daniil Yefimov Oct 18 '16 at 19:00
3

The argument regarding accumulation points doesn't make sense (what is $2.4$ here?) and is false, anyway: since $|\sqrt{2} - 1| < 1,$ the sequence of powers of $\sqrt{2} - 1$ tends to $0$.

The set is indeed neither open nor closed (in fact, it is dense in $\mathbb{R}$ since it is a ring containing numbers of arbitrarily small absolute value).

user378953
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  • Can I ask, why do we choose exactly this? $|\sqrt{2} - 1| $ And as i understood we need to show, that sequence of this number tends to zero, because we can take this number as our epsilon? I didn't get it – Daniil Yefimov Oct 18 '16 at 15:31