A wrong argument for $\mathbb{R}$ being countable

Question

We assume $A$ is the set of all countable subsets of the set of real numbers. We know $A$ is a partially ordered set $(A, \subseteq)$.

Suppose $$A_1 \subseteq A_2 \subseteq \ldots \subseteq A_n \subseteq A_{n+1} \subseteq \ldots$$ is a chain in $A$. We can prove $B=\bigcup_{n \in \Bbb{N}} A_n$ is a countable set.

For each natural number $m$, we have $A_m \subseteq B$. So $B$ is an upper bound for $A$. This shows each chain in $A$ has an upper bound according to Zorn's lemma. $A$ has a maximal element $X$, and we know $X$ is a countable set. Now we prove $X = \Bbb{R}$.

If $X \neq \Bbb{R}$, then there is an $x \in \Bbb{R}$ such that $x \notin X$. Let $Y=X \cup \{x\}$. It's obvious that $Y$ is a countable subset of the real numbers and $X \subsetneq Y$. This contradicts $X$ being a maximal element.

Thus, $X = \Bbb{R}$ and $\Bbb{R}$ is a countable set.

What is wrong with this argument?

Zorn’s Lemma requires every chain to have a maximal element. You only have that every chain of a certain sort has a maximal element. A chain is any totally ordered subset of a poset, not just countable ones. — Thomas Andrews, Jun 28 '21 at 19:55

Mike Earnest · Answer 1 · 2021-06-28T20:10:51.623

3

In order to use Zorn's lemma, you need to show that every chain has an upper bound. You have only shown that countable chains of the form $E_1\subseteq E_2\subseteq E_3\subseteq \cdots$ have an upper bound.

I cannot think of a constructive example of a chain in $A$ with no upper bound, but there is a nonconstructive one. Let $f:\omega_1\to \mathbb R$ be an injective function, where $\omega_1$ is the first uncountable ordinal number, equal to the set of countable ordinal numbers. Then the collection $\{f(A):A\in \omega_1\}$ is a chain, whose upper bound has cardinality $|f(\omega_1)|=|\omega_1|>\aleph_0$. Essentially, $f$ is a well-ordered, uncountable sequence of real numbers where every initial segment is countable.

Since $|\omega_1|\le |\Bbb R|$, such an injection $f$ exists, but I do not think we can construct such an $f$.

edited Jun 28 '21 at 20:10

answered Jun 28 '21 at 19:57

Mike Earnest

75,930

2

Isn't $\Bbb{Q}$ an upper bound of this chain? – Theo Bendit Jun 28 '21 at 19:58
@TheoBendit Yep! I removed the bad example. – Mike Earnest Jun 28 '21 at 20:00
1

I've been thinking, ever since I edited this question, of a good example, and I am coming up blank every time. There must be an example, or else $\Bbb{R}$ would indeed be countable! – Theo Bendit Jun 28 '21 at 20:01
1

you can get an example by well-ordering $\mathbb R$ – Asinomás Jun 28 '21 at 20:01
Yeah, it might not be possible to find a constructive example of a total ordering of countable subsets of $\mathbb R$ which doesn’t have a countable union. It is a bit like finding an uncountable ordinal. – Thomas Andrews Jun 28 '21 at 20:02
@Yorch You should write an answer. Even just a hand-wavy account of how to construct a chain without an upper bound would be a good addition to the question. – Theo Bendit Jun 28 '21 at 20:06
1

@TheoBendit I think the following works: https://math.stackexchange.com/a/4185566/33907 – Asinomás Jun 28 '21 at 20:26

A wrong argument for $\mathbb{R}$ being countable

1 Answers1