Is $\{e^{2\sqrt{2}k\pi i }|k\in\mathbb{Z} \}$ closed? And do these points all lie on a circle? This looks simple but for some reason, I cannot figure out the answers. Thank you
Asked
Active
Viewed 89 times
0
-
1You must have tried something? – Jun 18 '17 at 14:49
-
@Math_QED First, I thought this being countable means it is closed. But it is clearly wrong – Heisenberg Jun 18 '17 at 14:51
-
This question could help you : https://math.stackexchange.com/questions/1569152/dense-set-in-the-unit-circle-reference-needed?noredirect=1&lq=1 – Arnaud D. Jun 18 '17 at 14:52
-
@peterag Thank you for pointing it out. I edited the question – Heisenberg Jun 18 '17 at 14:52
-
@Heisenberg After this edition, it is trivial that your set is $S^1$, which is closed. Is this really what you meant? – José Carlos Santos Jun 18 '17 at 14:53
-
@JoséCarlosSantos Oh wow I messed this up. Let me edit this. This is not what I meant – Heisenberg Jun 18 '17 at 14:54
-
I edited the question again. There should not have been a t in the first place – Heisenberg Jun 18 '17 at 14:55
3 Answers
2
The (bilateral) sequence $\{e^{2\sqrt{2}\pi k i}\}_{k\in\mathbb{Z}}$ is dense in the unit circle since $\sqrt{2}$ is irrational.
If the set of elements of such sequence were closed, it would be the whole unit circle, but no element of the sequence equals $-1$, for instance.
Jack D'Aurizio
- 353,855
2
Every point of your set belongs to $S^1$. Since $\sqrt2\pi$ irrational, it follows from Dirichlet's approximation theorem that it is a dense subset of $S^1$. Therefore, it is not closed.
José Carlos Santos
- 427,504
1
The points indeed lie on the circle within the complex plane, since $|e^{\theta i}| = 1$ for any $\theta \in \Bbb R$.
The set is not closed. In fact, we can show that the closure $\overline{\{e^{2\sqrt{2}k\pi i }|k\in\mathbb{Z} \}}$ is the entire unit circle with an argument along these lines. That is, we want to show that $\{\langle \sqrt{2}\,m \rangle : m \in \Bbb Z\}$ is dense in the interval $[0,1]$.
Ben Grossmann
- 225,327