# Continuous functions on $\omega_1$

Let $\lambda<\omega_1$ be a limit ordinal and consider the set $X=\{\alpha+1\colon \alpha<\lambda\mbox{ is a limit ordinal}\}$ Is the characteristic function $\chi_X$ continuous on $\omega_1$ with interval topology?

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Notice that $\omega \cdot \omega = \sup_{n\in\omega} \omega \cdot n = \sup_{n\in\omega} (\omega\cdot n+1)$.

This implies that in the interval topology $\lim_{n\to\infty} (\omega \cdot n+1)= \omega\cdot\omega$. But $1=\lim_{n\to\infty} f(\omega\cdot n+1)\ne f(\omega\cdot\omega)=0$, and thus this function is not continuous.

I assume that by interval topology you mean the topology generated by the subbase consisting of intervals $\{\xi; \xi<\beta\}$ and $\{\xi; \beta<\xi<\omega_1\}$ for $\beta<\omega_1$, see e.g Komjath, Totik: Problems and theorems in classical set theory p.40. Note that this book also contains an exercise containing some characterization of functions which are continuous with respect to interval topology.

Martin: in the second limit equation you forgot $f$. Why do you use \ldot instead of \cdot? – Damian Sobota Nov 29 '11 at 23:18
@Damian Thanks for noticing the forgotten $f$. I am used to write $a.b$ or $ab$ instead of $a\cdot b$. I believe it is not that unusual. According to wikipedia Multiplication is sometimes denoted by either a middle dot or a period: $5\cdot 2$ or $5.2$. Although of course my notation is confusing for people from countries where period is used as decimal separator. – Martin Sleziak Nov 30 '11 at 12:19
In Britain they prefer $a.b$ for multiplication and $3\cdot 14159$ for decimals. The copy editor for the journal will even change your manuscript for you to their system! – GEdgar Nov 30 '11 at 15:14