6
$\begingroup$

Is $$ \bigcup_{n=1}^{\infty}\{z\in\Bbb C:z^n=1\}=\{z\in\Bbb C:|z|=1\}$$my argument is that the argument of the elements of the first set are rational multiples of $\pi$ whereas the second set also consists of elements with irrational multiples of $\pi$. But I am not so sure, there is still some doubt.

$\endgroup$
17
$\begingroup$

Those two sets are certainly not equal.

$\displaystyle\bigcup_{n=1}^{\infty}\{z\,|\,z^n=1,n\in \mathbb N\}$ is countable as it is a countable union of finite sets, while $\{z/|z|=1\}$ has the cardinality of the continuum.

$\endgroup$
  • 1
    $\begingroup$ What does "power of a continuum" mean? $\endgroup$ – Karthik Feb 1 at 15:21
  • 4
    $\begingroup$ I edited as power of the continuum is a too fast translation from French! I mean the cardinality of the continuum, i.e. has the same cardinality as $\mathbb R$. $\endgroup$ – mathcounterexamples.net Feb 1 at 15:23
  • 1
    $\begingroup$ "power of the continuum" is also a common term for it in English, but is perhaps becoming antiquated (as am I). $\endgroup$ – Paul Sinclair Feb 1 at 17:22
  • 1
    $\begingroup$ @PaulSinclair I'm afraid of becoming antiquated too. Let's reassure ourselves in saying that we're like the good Armagnac's (or Cognac's depending on your taste). $\endgroup$ – mathcounterexamples.net Feb 1 at 17:35
  • $\begingroup$ This is correct, but feels like a sledgehammer to crack a nut. Maybe why it's obvious the LHS contains only rational powers of $i$ would be easier to follow. $\endgroup$ – drjpizzle Feb 1 at 18:22
11
$\begingroup$

Your argument is correct.

For example, $e^{\sqrt{2} \pi i}$ is not in the first set.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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