0
$\begingroup$

In the Wiki of order topology, I encounter the following statement.

$\omega_1$ is a limit point of the subset $[0,\omega_1)$ even though no sequence of elements in $[0,\omega_1)$ has the element $\omega_1$ as its limit.

I have no idea how to prove it.


Recall that for any topological space $X$ and a subset $Y$ in $X,$ we say that $y\in Y$ is a limit point of $Y$ if every neighbourhood of $y$ contains at least one element of $Y$ different from $y$ itself.

$\endgroup$
  • $\begingroup$ In which topological space are you working? $\endgroup$ – Lord Shark the Unknown Aug 22 '18 at 12:10
  • $\begingroup$ I am working in ordinal space. $\endgroup$ – Idonknow Aug 22 '18 at 12:11
  • 2
    $\begingroup$ What is "ordinal space"? (don't forget Burali-Forti!) $\endgroup$ – Lord Shark the Unknown Aug 22 '18 at 12:12
  • $\begingroup$ Set of ordinal numbers with order topology $\endgroup$ – Idonknow Aug 22 '18 at 12:13
  • $\begingroup$ Burali-Forti${}$! $\endgroup$ – Lord Shark the Unknown Aug 22 '18 at 12:16
3
$\begingroup$

Your statement splits into two.

$\omega_1$ is a limit point of $[0, \omega_1)$.

Indeed, suppose that we have an open $(a, b)$ that contains $\omega_1$ (after all, open intervals, together with rays, form a topology base). Then $a < \omega_1$, so $a$ is countable, and the successor ordinal will be as well, proving the statement.

No sequence within $[0, \omega_1)$ converges to $\omega_1$.

Let $(x_n)_{n \in \mathbb N}$ be any sequence in $[0, \omega_1)$. Then their union $x_\infty$ will be a countable upper bound, since the union of countable sets is countable. Thus, $(x_\infty, \omega_1 + 1)$ will be an open set containing $\omega_1$ but no element of $(x_n)_{n \in \mathbb N}$.

$\endgroup$

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.