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I'm trying to understand the following. I'll denote by $X$ the set of all ordinals that are at most $\omega_1$, the first uncountable ordinal. To avoid confusion, they can be equal to $\omega_1$ as well. Equip $X$ with the order topology.

Then how come $\omega_1$ is an accumulation point of $X\setminus\{\omega_1\}$, but there is no sequence in $X\setminus\{\omega_1\}$ that converges to $\omega_1$?

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@AsafKaragila: Both links are the same. – Damian Sobota Dec 13 '11 at 20:08
Sorry, how does not being first countable give the result? – Dr. Zeus Dec 13 '11 at 20:11
See this, and this. In short, $\omega_1$ is not first countable. – Asaf Karagila Dec 13 '11 at 20:18
@Damian: I corrected that, thanks. – Asaf Karagila Dec 13 '11 at 20:18
@Dr.Zeus: If the space is not first countable then sequences are not "enough" to determine all the limits. Topological spaces are much more general than you'd expect. – Asaf Karagila Dec 13 '11 at 20:23
up vote 6 down vote accepted

Take a neighbourhood $U$ of $\omega_1$. By the definition of order topology $U$ contains an open interval $(\alpha,\omega_1]$ for some $\alpha<\omega_1$. Note that $\alpha+1<\omega_1$ so that $\alpha+1\in U\cap(X\setminus\{\omega_1\})$. This shows $\omega_1$ is an accumulation point of $X\setminus\{\omega_1\}$.

Take a sequence $(\alpha_n)_{n\in\mathbb{N}}$ in $X\setminus\{\omega_1\}$. Consider the ordinal $\beta=\bigcup_{n\in\mathbb{N}}\alpha_n$. As a countable union of countable sets it's still countable, so $\beta\in X\setminus\{\omega_1\}$. Now $(\beta,\omega_1]$ is a neighbourhood of $\omega_1$ containing no elements of the sequence $(\alpha_n)$, so $(\alpha_n)$ can't converge to $\omega_1$. This shows no sequence of $X\setminus\{\omega_1\}$ converges to $\omega_1$.

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