Why does the set of countable sets of $\mathbb R$ does not satisfy the conditions of Zorn's lemma? In the subject of set theory a student says the following claim :
'' Exists at least one maximal countable subset of $\mathbb{R}$.That it's true because the set of countable subsets of $\mathbb{R}$ is partially ordered set with order, the relation of 'containing'.Every chain of these sets has supremum (the union of chain).So from the Zorn's Lemma the set has maximal element.''
That proof seems right and i can't find false.Can you find any false on this claim?
 A: Let $\preceq$ be a well-ordering of $\Bbb R$. For each $x\in\Bbb R$ let $P_x=\{y\in\Bbb R:y\prec x\}$. $\Bbb R$ is uncountable, and there are two possibilities.


*

*If $\{x\in\Bbb R:P_x\text{ is uncountable}\}\ne\varnothing\}$, let $$x_0=\min\{x\in\Bbb R:P_x\text{ is uncountable}\}\;;$$ then $P_{x_0}$ is uncountable, but $\{P_x:x\prec x_0\}$ is a chain of countable subsets of $\Bbb R$ whose union is $P_{x_0}$.

*Otherwise, $\{P_x:x\in\Bbb R\}$ is a chain of countable subsets of $\Bbb R$ whose union is $P_{x_0}$.

A: That assumes that the union of a chain of countable sets is countable too. That is not necesarily true, although it is true if the chain happens to be countable.
A: Your mistake, as remarked by other answers already, is that some chains are not countable, and their union is not countable either.
But there is a deeper issue at play here. People forget that the supremum, or indeed just upper bound, needs to exist within the partial order.
It is easy to say that a chain of subsets has an upper bound: its union, but it's not easy to prove that this union is in your partial order. And why would it be? We know there is no maximal countable subset. Since the reals are uncountable, any countable subset is not everything, so can be extended by at least one more element.
Much like how $[0,1)$ has an upper bound in $[0,1]$, but not in $[0,1)$. 
