Suppose we have a set $M = (0,1) \subset R$ of reals well-ordered as the first uncountable ordinal.


Let $M(a) = \lbrace x \in M : x < a \rbrace$. For every $a \in M$ set $M(a)$ is countable. That's why every increasing sequence is bounded:  

$$(\*) ~~~~~~~~~~ \forall \lbrace a\_1,...,a\_n,...\rbrace \subset M ~~\exists b \in M : a\_i < b ~~\forall i \in \mathbb{N}.$$


Now suppose that we can pick elements from $M$ at **random**. And let's try to build an increasing random sequence by the following algorithm. Let's say we have an increasing sequence of elements $\{a\_1,...,a\_n\}$. Pick some random number $b$. If $b > a\_n$ set $a\_{n+1} = b$. Otherwise pick other random number instead of $b$ and check $b > a\_n$ condition. Continue this till success. Since we pick numbers at random it shouldn't be a problem to construct an infinite sequence. But every infinite sequence is bounded! Which means that in our infinite process we will never be able to pick random number which is greater than some number $c \in M$. This is even more astonishing since $M(c)$ is countable and $M \setminus M(c)$ is uncountable!


Any thoughts how to "solve" this paradox?



The paradox with the first uncountable ordinal
I think the paradox can be solved by the observation that $c$ is defined a posteriori, so it is not true that you avoid $c$ during all the process, since $c$ is in fact an outcome of the process.


You could do a similar paradox without bothering with well-orderings: pick a sequence $X=\{a\_1,a\_2,\dots\}$ in $(0,1)$ by choosing each $a\_i$ at random, one after the other.


Then your "paradox" is that each $a\_i$ you chose belong to $X$, which has measure $0$ and is countable, whereas $(0,1)$ is uncountable with measure $1$. But since $X$ is the outcome of the process and was not defined at the time of the choosing, it is not a real paradox.


