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.

  • $\begingroup$ 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

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 $(x_n)_{n \in \mathbb N}$.


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