Confused about Cantor's diagonal argument

Question

If I enumerate the even natural numbers (E), the enumeration won't contain the number 3. It seems to me that this proves that E forms a proper subset of the natural numbers (N). But that does not prove that E cannot be enumerated, or that E cannot be bijected onto N, or that N cannot be enumerated.

Cantor's diagonal argument demonstrates that his list of real numbers (L) does not contain a particular real number. Surely that proves that L is a proper subset of the real numbers (R). But how does it prove that L cannot be enumerated, or that L cannot be bijected onto R, or that R cannot be enumerated?

Answer 1

Cantor's diagonal argument works on any proposed enumeration $L$, not just a particular proposal (which might fail for a more obvious reason). Basically any proposed enumeration of the reals must be missing something; that's what his argument says.

Answer 2

I find that it helps to recast Cantor's argument as constructing something.

Specifically, we define a function $F$ from $\{$maps from $\mathbb{N}$ to $\mathbb{R}\}$ to $\mathbb{R}$, that is, from lists of reals to reals. This $F$, in particular, "does something" to every list.

We now prove a fact about the $F$ we've just constructed: namely, that - regardless of what list $f$ we start with - $F(f)$ is not in $ran(f)$.

We have now used the object $F$ we've constructed to show that no counting of $\mathbb{R}$ exists. But this depends crucially on the part of the previous paragraph that says

regardless of what list $f$ we start with

and it's here that your idea breaks down. What you've shown is that there is some list of naturals which is not surjective; what you would need to do is show that every list of naturals fails to be surjective.

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EDIT: Let me add the following, which I hope will help.

You are quite right that "is a proper subset of" does not imply "is smaller than". So just knowing that $ran(f)\not=\mathbb{R}$ for some listing $f$ of reals does not mean that $ran(f)$ is smaller than $\mathbb{R}$!

However, again, note that Cantor has shown that every $f:\mathbb{N}\rightarrow\mathbb{R}$ has the property that $ran(f)\not=\mathbb{R}$. It is the fact that every $f$ has "small range," not the fact that some $f$ has "small range," which proves that $\mathbb{R}$ is uncountable.

One way to think about this that might help is the following. Suppose $f:\mathbb{N}\rightarrow\mathbb{R}$. By Cantor's argument, we know $ran(f)\not=\mathbb{R}$. But couldn't $ran(f)$ have the same size as $\mathbb{R}$, nonetheless?

Well, suppose it did - that is, suppose there were a surjection $g$ from $ran(f)$ to $\mathbb{R}$. Then think about $g\circ f$. This is a function from $\mathbb{N}$ to $\mathbb{R}$, and - by assumption on $g$ - it is surjective.

But that's a problem! Cantor's argument shows that no map from $\mathbb{N}$ to $\mathbb{R}$ can be surjective. So we have a contradiction.

So not only is $ran(f)$ a proper subset of $\mathbb{R}$, for every $f:\mathbb{N}\rightarrow\mathbb{R}$, but $ran(f)$ is always strictly smaller than $\mathbb{R}$!

This is an interesting phenomenon, that happens a lot in mathematics: a weak universal statement can sometimes "strengthen itself." Basically, what this looks like is the following:

• We prove "Every Foo has the weak Bar property."

• Now, in general, the weak Bar property does not imply the strong Bar property.

• However, from a Foo without the weak Bar property, we can build a Foo without the strong Bar property.

• So by the first bullet point, we've in fact shown: Every Foo has the strong Bar property.

(The following paragraph is subjective, and you may not find it helpful - feel free to ignore it if so. For contrast, consider the phenomenon of strengthening the induction hypothesis in proofs by induction: you want to prove "All $n$ have property $\alpha$" by induction, but the obvious attack doesn't work - so instead you prove "All $n$ have property $\beta$" for a stronger property $\beta$! And, even though this should be harder to prove (you're proving a stronger result after all), it winds up being easier. That is, to prove a weak thing, you find yourself having to prove a strong thing. What's going on above is sort of the dual of this: we prove a weak thing, and then that immediately implies the strong thing!)

Answer 3

Cantor's diagonal argument does not "demonstrate that his list of real numbers (L) does not contain a particular real number", it proves that for any list of real numbers, there always exists a real number not on the list. Your fact about the list of even numbers does not make the same assertion - you construct a natural number not on a specific list, Cantor constructs a real number not on an arbitrary list.

Answer 4

The structure of Cantor's argument is:

Suppose $L$ is a foo.

Then (insert some derivations here), ergo contradiction.

Ergo, there are no foos.

In particular:

Suppose $L$ is a surjection $\mathbb{N} \rightarrow \mathbb{R}$.

Then (insert some derivations here), ergo contradiction.

Ergo, there are no surjections $\mathbb{N} \rightarrow \mathbb{R}$.

The idea behind this style of argument is really that by dealing with an arbitrary $L$, it's kind of like dealing with every possible $L$ at once. We conclude that every possible $L$ leads to a contradiction, and therefore that there are no $L$'s with the property of interest.