Where is the mistake in the argument in favor of the (erroneous) claim "every Dedekind cut is a rational cut"?

Question

A cut is a set $C$ such that:

(a) $C\subseteq \mathbb Q $

(b) $C \neq \emptyset $

(c) $C \neq \mathbb {Q} $

(d) for all $a, c \in \mathbb Q $ , if $c\in C$ and $a\lt c$ , then $a\in C $

(e) for all $c\in \mathbb Q$ , if $c\in C$, then there exists $d\gt c$ such that $d\in C$

A rational cut is a cut of the form $C(r) = \{c | c\lt r, r\in \mathbb Q\}$.

I know that the statement "every cut is a rational cut" can be disproved by giving a counterexample. However, I can't prevent myself from finding an air of plausibility to the following argument aiming at justifying that every cut is of the form $C(r)$.

Hence my request: can you please show me where the following argument goes wrong?

Step 1 :

Suppose $C$ is a cut , and that $C$ is not of the form $C(r)$ for some $r \in \mathbb Q$.

This amounts to saying:

(1) there is no $r$ such that for all $c$ if $c \in C$ then $c\lt r $

(2) implying that: for all $r$ it is not the case that for all $c$ if $c \in C$ then $c\lt r $

(3) implying that: for all $ r$ there is some $c$ such that $c\in C$ and $c\geq r $

Step 2:

Suppose $r\in \mathbb Q$ and $r\notin C $.

By (3) above there is some $c$ such that: $c\in C$ and $c\geq r$.

Now, if $c=r$ then $r\in C$.

Also, if $c\gt r$ then $ r\lt c$ and, by the "cut" definition, $r\in C$.

So the hypothesis $r \notin C$ leads to a contradiction; meaning that $\mathbb Q\subseteq C$, since $r$ was arbitrary.

Finally we have $C\subset \mathbb Q$ and $\mathbb{Q} \subseteq C$, a contradiction (since the first statement negates that all rationals belong to $C$ while the second asserts the same proposition).

The conclusion seems to be that a Dedekind cut that is not a rational cut is not a Dedekind cut; in other words, every Dedekind cut has to be a rational cut.

Note: I can see an objection to the conclusion of "step 2", namely that what the argument shows is that the first conjuction is false, namely $r\in \mathbb Q$. But how to make sense of the negation of this conjunction since the set of rationals is the domain over which range all our number variables (in this argument)?

Answer 1

(1) in Step 1 contains an error: the "if" in (1) [and (2)] should be "if and only if", which wouldn't allow to conclude (3).

The corrected (3) becomes:

for all $r$, there is some $c$ such that either $c\in C$ and $c\geq r$, or $c\notin C$ and $c<r$.

Answer 2

colt_browning has given an excellent answer already pinpointing the mistake, but here’s a general tip for finding such mistakes, in places like this where you think you’ve proven something that you know shouldn’t be true in general: Take a counterexample to the general statement, and follow through the proof checking each claim against that counterexample.

Here, for instance, you know that any irrational number should give a counterexample — for instance, the Dedekind cut $D_{\sqrt{2}} = \{ x \in \mathbb{Q}\ |\ x < \sqrt{2} \}$. Your proof starts with an arbitrary Dedekind cut, which could be $D_{\sqrt{2}}$, and ends up claiming that it’s rational, which $D_{\sqrt{2}}$ is not. So look at each claim, and check whether it holds for $D_{\sqrt{2}}$. This may be easier with some statements than others, depending how concrete they are; so you can start anywhere in the middle, with whichever claim seems easiest to check, and if that fails for $D_{\sqrt{2}}$, you know an error has already occurred by that point.

In particular, in your case, I scanned through and (3) — “for all $r \in \mathbb{Q}$, there is some $c \in C$ with $c \geq r$” — stands out as false, for $D_{\sqrt{2}}$ (and, once one notices this, for any cut) — it says that the cut is unbounded above in the rationals, which certainly isn’t right. So the error must be before (3) — and checking carefully back, you can come to the specific error colt_browning’s answer points out: that in (1), the characterising property of a rational cut should involve an “if and only if”, not just an “if”.

Answer 3

=== new better answer ====

Being a rational cut of the form $C = \{c| c < r\}$ [assuming our "known universe' is only the rational numbers] means not only that $r$ is an upper bound, but that $r$ is the least upper bound.

Not only does it mean that if $c \in C$ then $c < r$. It also means that if $c \in C$ then $c \ge r$. And it also means that if $q < r$ then there is a $c \in C$ so that $q < c$.

$r$ is, of course, not the *only $r > $ all $c \in C$. After all, if $c \in C$ then $c < r < r+1 < r + 3.5 < r + 17< ....$. But what is special about $r$ is that if $c \not \in C$ then $c \ge r$. That is obviously not true for $r+1, r+3.5$ etc. Also if $q < r$ then there is a $c\in C$ so that $q < c < r$. That is obviously not true for $r+1, r+3.5$ etc.

A cut that isn't rational means that there is no rational $r$ that acts as a least upper bound. There will always by upper bounds no upper bound will by least.

So you statement 1,2,3 are all incorrect.

They should read:

1. There is no least $r$ so that if $c \in C, c < r$. If $c \in C$ implies $c < r$ then there is an $r' < r$ so that if $c \in C$ then $c < r' < r$.

2. for all $r$ it is not the case that $c \in C \iff c < r$. For all $r$ there will either be $c\in C$ where $c \ge r$ OR there will be a $c < r$ where $c\not \in C$.

3. for all $r$ there is either a $c \in C$ where $c \ge r$ or there is a $c \not \in C$ were $c < r$.

.... from here on.... your argument just won't work.

=== old answer ====

"Suppose C is a cut , and that C is not of the form C(r) for some r∈Q

.

This amounts to saying:

(1) there is no r such that for all c if c∈C then c<r"

Being of the form $C= \{c\in \mathbb Q| c < r\}$ means that 1) If $c \in C$ then $c < r$ but it ALSO means 2) If $c \not \in C$ then $c \ge r$.

There will always be an $r$ so that if $c \in C$ then $c < r$. For example if we are talking about the cut for $\pi$ then if $c \in C$ then $c < 27439827583942798$ and $ 27439827583942798$ is certainly rational.

But $c\not \in $ the cut for $\pi$ does not mean that $c \ge 27439827583942798$. for example $c = 3.5$ is not in the cut for $\pi$ but $3.5 < 27439827583942798$.

What we can say is there is not $r$ where $c\in C \iff c < r$. For any rational $r$ then either $r < \pi$ or $r > \pi$. If $r < \pi$ then there are $c \in C$ so that $r < c$. And if $r > \pi$ there is a $c \not \in C$ so that $c < r$.

From here step 2 "implying that: for all r it is not the case that for all c if c∈C then c<r" needs to be modified to " For all $r\in \mathbb Q$ it is not the case that if $c\in C$ then $c < r$ and if $c \not \in C$ then $c \ge r$".

And 3) need to become "for all $r \in \mathbb Q$ there is either a $c\in C$ where $c \ge r$ OR there is a $c\not \in C$ where $c < r$"

And from there, no contradiction arises.