I wanted to {$e^{ik}|k\in N$} Is dense in $S^1$
My Attempt:
On the contrary, assume that it is not dense in $S^1$.
So there exists a point in $S^1$ which is not a limit point, say $e^it $ for some $t\in R$
That means $\exists \epsilon >0$ such that $\forall k\in N ,|e^{ik}-e^{it}|>\epsilon $ that is
$|e^{i(k-t)}-1|>\epsilon $
I was expecting some contradiction to occur.
But I am not able to see.
Any help will be appreciated
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Bernard
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Curious student
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1Perhaps you could try to show that $k \mod 2 \pi$ is dense in $[0,2 \pi]$? – copper.hat Sep 04 '18 at 15:19
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1Your proof needs to use that $2\pi$ is irrational. – Kusma Sep 04 '18 at 15:20
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That is an irrational statement. – copper.hat Sep 04 '18 at 15:23
1 Answers
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As already in the comments indicated, you have to use that $\pi$ is irrational. The key tool here is Kronecker's Approximation Theorem. The proof of Kronecker's Approximation Theorem is quite simple (it is an application of the pigeonhole-principle), but there are no 'simple' proofs of the irrationality of $\pi$, see here for more details.
p4sch
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