# R is countable using Zorn's Lemma?

I used Zorn's Lemma to cook up a proof that $\mathbb{R}$ is countable, and now I can't find a flaw in it. Can anyone help? Here it is...

Denote by $\mathcal{A}$ the set of countable subsets of $\mathbb{R}$, partially ordered under inclusion, and consider any chain in $\mathcal{A}$: $$A_0 \subseteq A_1 \subseteq A_2 \subseteq \dots$$ Now let $A = \cup_{k \ge 0} A_k$. Then $A \in \mathcal{A}$, the countable union of countable sets.

It follows that every chain in $\mathcal{A}$ has an upper bound. So by Zorn's Lemma, $\mathcal{A}$ contains an element $X$ maximal under inclusion.

Suppose now that there is $a \in \mathbb{R}$ not in $X$. Then $X' = X \cup \{ a \}$ is countable, for if $f: X \to \mathbb{Z}^{+}$ is a bijection, then $f': X' \to \mathbb{Z}^{+}$, $$f'(a) = 1 \quad \text{and} \quad f'(x) = f(x) + 1,$$ is a bijection. But then $X' \in \mathcal{A}$ strictly contains the maximal element $X$, a contradiction. So $a \in X$, and hence $X = \mathbb{R} \in \mathcal{A}$.

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Line 6 assumes that the chain is countable. A chain of countable sets needs not be countable in length, and its union does not need to be countable. See this. – Andrés E. Caicedo Mar 12 '14 at 5:50
Fantastic, thanks. If you add it as an answer I'll definitely accept it. – Joshua Ciappara Mar 12 '14 at 6:19

Line 6 assumes that the chain is countable. A chain of countable sets needs not be countable in length, and its union does not need to be countable. For instance, $\omega_1$, the first uncountable ordinal, is the (well-ordered) union of a chain of countable sets.