9
$\begingroup$

Let $(x_{n})_{n \geq 1}$ sequence in $\mathbb{R}$. Is there a sequence with an uncountable number of accumulation points?

Thank you!

$\endgroup$
  • $\begingroup$ if sequence itself is countable, how can it have an uncountable accumulation point? $\endgroup$ – gt6989b Apr 4 '16 at 0:28
  • 4
    $\begingroup$ Well a sequence can have uncountably many limit points. The set of limit points could even be the entire real line. $\endgroup$ – Matt Samuel Apr 4 '16 at 0:29
  • 1
    $\begingroup$ Easily enough, @gt6989b actually. $\endgroup$ – Thomas Andrews Apr 4 '16 at 0:29
  • 3
    $\begingroup$ @gt6989b I see what you are thinking. You are probably saying "My sequence is countable, so the number of subsequences is countable, so the number of limit points is countable". The number of subsequences of a sequence isn't actually countable. Use diagonalization for the proof (write all subsequences one below the other, and use the diagonal to form a new subsequence that doesn't match with any of the written subsequences). Hence the number of limit points need not be countable, as is the case with rational numbers. $\endgroup$ – астон вілла олоф мэллбэрг Apr 4 '16 at 0:38
  • 4
    $\begingroup$ Fun fact: in a more general Hausdorff space, the number of cluster points can even be more than $2^{\aleph_0}$, which is the number of subsequences. $\endgroup$ – Nate Eldredge Apr 4 '16 at 4:01
23
$\begingroup$

Since the rational numbers are countable, we know there is a bijection $f:\mathbb N\to\mathbb Q$. Let $x_n=f(n)$. Then this is a sequence which contains every rational number, and its set of accumulation points is $\mathbb R$.

$\endgroup$
  • 4
    $\begingroup$ For example $\frac12, \frac13,\frac23, \frac14,\frac24,\frac34, \frac15,\frac25,\frac35,\frac45, \frac16,\frac26,\ldots$ is a sequence whose set of accumulation points is $[0,1]$ which is of course uncountable. $\endgroup$ – Jeppe Stig Nielsen Apr 4 '16 at 11:59
5
$\begingroup$

Terms of a sequence $$(\sin n)_{n\in\mathbb N}$$ are dense in $[-1,1]$ so each number in the interval is an accumulation point of the sequence.

$\endgroup$

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.