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Some things I know:
$\omega_1$ is the first uncountable ordinal, made up of all of the countable ordinals
$f$ is not necessarily continuous

For this proof, I currently have $x$ as a condensation point in my range, and I know that $(x - 1/n , x + 1/n )$ is uncountable and everything to the left and right of this interval is countable. I understand that I want to shrink this interval small enough just to $x$ (which will be my constant), making everything around $x$ eventually countable. I also understand that this will eventually make all of the elements in $\omega_1$ go to $x$ but I am having a hard time seeing where my contradiction comes in.
I also am struggling with writing this proof in a presentable fashion.

Thank you!

Lucenaposition
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L. Clemente
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2 Answers2

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HINT: There is an order-isomorphism $h:\Bbb R\to(0,1)$. If you had a strictly increasing $f:\omega_1\to\Bbb R$, you could compose it with $h$ to get a strictly increasing function from $\omega_1$ to $(0,1)$, so there is no harm in assuming that $f$ has bounded range. Let $a=\sup_{\xi\in\omega_1}f(\xi)$, and for $n\in\omega$ let

$$A_n=\{\xi\in\omega_1:f(\xi)<a-2^{-n}\}\;.$$

  • What can you say about the cardinalities of the sets $A_n$?
  • Now consider $\bigcup_{n\in\omega}A_n$.

A completely different approach is to show that every uncountable $X\subseteq\Bbb R$ contains a two-sided condensation point, i.e., a point $x\in X$ such that every set of the form $(a,x]$ with $a<x$ and every set of the form $[x,a)$ with $x<a$ contains uncountably many points of $X$. You can find a proof here.

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Brian M. Scott
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  • For the two-sided condensation point approach, the whole interval (x - (1/n) , x + (1/n) ) is uncountable in a Lindelof space. Therefore it has a countable sub cover, say C. And this is a contradiction? – L. Clemente Nov 06 '16 at 19:51
  • @L.Clemente: No, that is not what I was suggesting. The point is that if $X={f(\xi):\xi\in\omega_1}$, there is a $\xi\in\omega_1$ such that $(\leftarrow,f(\xi)]$ is uncountable, and that is a contradiction. – Brian M. Scott Nov 06 '16 at 19:58
  • And that is a contradiction because everything outside the interval is countable? – L. Clemente Nov 06 '16 at 20:17
  • @L.Clemente: No, it’s a contradiction because it implies that $\xi$ has uncountably many predecessors in $\omega_1$. – Brian M. Scott Nov 06 '16 at 20:20
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    A slight variant would start with the observation that it's impossible for $f$ to be unbounded; otherwise, if you choose $\alpha_n < \omega_1$ such that $f(\alpha_n) > n$, and let $\beta = \sup { \alpha_n } < \omega_1$, then $f(\beta) > n$ for every $n$ giving a contradiction. – Daniel Schepler May 06 '19 at 23:53
  • I understand that $|A_n|$ is supposed to be countable, but I'm not sure I see why that's the case. Can you explain? – aduh Nov 18 '20 at 04:03
  • @aduh: $a-2^{-n}<a=\sup_{\xi\in\omega_1}f(\xi)$, so there is an $\eta<\omega_1$ such that $a-2^{-n}<f(\eta)<a$. $|A_n|\le|\eta|\le\omega$. – Brian M. Scott Nov 18 '20 at 04:07
  • Ah, right, I see. Thanks for the quick reply! – aduh Nov 18 '20 at 04:09
  • @aduh: You’re welcome! – Brian M. Scott Nov 18 '20 at 04:11
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If you are having a hard time proving that if $f$ is not eventually constant, then it has a a strictly increasing subsequence, consider the following:

Let $\alpha$ be an ordinal, let $(X,<)$ be an ordered set, and let $f: \alpha \rightarrow X$ be increasing. Either $f$ is eventually constant, or there is a strictly increasing and cofinal map $\varphi: cof(\alpha) \rightarrow \alpha$ such that $f \circ \varphi$ is strictly increasing.

Where $cof(\alpha)$, namely the cofinality of $\alpha$, is the least ordinal $\beta$ such that there is a cofinal map $\beta \rightarrow \alpha$.

I leave it to you to prove this if you like (one can define $\varphi$ inductively on $cof(\alpha)$).

Then, prove that $cof(\omega_1) = \omega_1$ and you get a strictly increasing subsequence.

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