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Although I have read that it's quite easy to prove there isn't an $\omega_1$ increasing sequence on real set I spent a lot of time figuring out why it happens and finally I think I made it, but I'm not sure about it. Here is my approach.

First of all, I supossed there was an increasing sequence $s: \omega_1\longrightarrow\mathbb{R}$ and I considered the following sets

$D_n=\{\alpha<\omega_1\;\vert\;s(\alpha+1)-s(\alpha)<\frac{1}{n}\}$

After that, to demonstrate that there was a set $ D_n $ such that its cardinality was $\aleph_1 $, I assumed not, but that would mean that all sets $ D_n $ were countable and then their union would also be countable so that there is $ \aleph_1 $ ordinals less than $ \omega_1 $ so that each of them does not belong to any set $ D_n $ (due to the regularity of $ \omega_1 $), that would mean that for all of them $ s (\alpha + 1 ) -s (\alpha) \geq 1 $ messing up the injection of $ s $ because there would be $ \aleph_1 $ ordinals in the domain and $ \aleph_0 $ in the range.

Finally, the existence of a $D_{n_{0}}$ set such that its cardinality is $\aleph_1$ is a contradiction for the following reason. Take a random $\alpha$ in $D_{n_{0}}$, then $s(\alpha+1)-s(\alpha)<\frac{1}{n_0}$ but also, by the archimedean postulate, there would be a natural number $m$ such that $\frac{1}{m}<s(\alpha+1)-s(\alpha)<\frac{1}{n_0}$ and that number $\frac{1}{m}$ would never reach $\frac{1}{n_0}$ ,because there are $\aleph_1$ numbers between them, and that would contradicts the archimedean postulate.

Am I right up to this point?

Thanks in advance for your help and time.

PD: I'm a begginer Set theory student so sorry if I said something that didn't make sense.

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I’m afraid that I don’t understand your argument that no $D_n$ can be uncountable; in particular, I can’t make any sense of the assertion that ‘$\frac1m$ would never reach $\frac1{n_0}$’.

In any case, you’re working much too hard: all you have to do is note that each interval $\big(s(\alpha),s(\alpha+1)\big)$ must contain a rational number $q_\alpha$, and since these intervals are pairwise disjoint, $\{q_\alpha:\alpha\in\omega_1\}$ would then be an uncountable set of rational numbers, which is absurd.

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  • $\begingroup$ Well, my idea was that if all the sets $D_n$ were countable I would only have $\omega$ ordinals in the union of them, but I have in total $\omega_1$ numbers so that for the another ordinals that do not belong to any $D_n$ set the difference between it and its successor would be at least 1. That would mean a contradiction? $\endgroup$ – Marinovsky Jun 28 '20 at 18:00
  • $\begingroup$ In the other hand, thanks for the reply I really don't know why I didn't figure out that. $\endgroup$ – Marinovsky Jun 28 '20 at 18:02
  • $\begingroup$ @Marinovsky: I understood why the $D_n$ cannot all be countable; what I don’t understand is your argument that none of them can be uncountable. And your original argument doesn’t even mention ordinals not belonging to any of them, though it would not be hard to show that that set must be countable. (And you’re very welcome.) $\endgroup$ – Brian M. Scott Jun 28 '20 at 18:09
  • $\begingroup$ Ohh, now I can see where is my mistake, thanks again for that Mr.@Scott. $\endgroup$ – Marinovsky Jun 28 '20 at 18:37
  • $\begingroup$ @Marinovsky: You’re welcome! $\endgroup$ – Brian M. Scott Jun 28 '20 at 18:39

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