Show an increasing function f from omega_1 to R is eventually constant 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!
 A: 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.
A: 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.
