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?
Continuous functions on $\omega_1$
1 Answers
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.
You can find out more about this book on the website of one of the authors.
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0Martin: in the second limit equation you forgot $f$. Why do you use \ldot instead of \cdot? – 2011-11-29
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0@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](http://en.wikipedia.org/wiki/Multiplication#Notation_and_terminology) _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](http://en.wikipedia.org/wiki/Decimal_separator). – 2011-11-30
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0In 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! – 2011-11-30