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Let us fix a well ordering of the real numbers then consider a 'list' of some subset of the real numbers (with at least two elements) - called A -, enumerated by the well ordering.

Say our well order is $ r_1, r_2 $ , ... [I know the real aren't countable].

Then to every 'sequence', $s$, (which itself is a function from the reals to the reals, where we start our sequence at $r_1$) let $s_{r_j}$ be the value of the sequence assigned to real number $r_j$ of the well ordering.

For example, let us say our well order looks like 1,5,6,1/23, at the start. Let our subset A be {2,5} Then the sequence (which is function from the real to the reals) with first 3 values 2 and fourth value 5 would be the function sending 1 and 5 and 6 to 2 and 1/23 to 5 - that is $s_1=s_{r_1} = 2$, $s_5=s_{r_2}=2$,$s_6=s_{r_3}=2$,$s_{1/23}=s_{r_4}=5$

Now, if the set of all such sequences had the same cardinalty as the real numbers, for ever sequence s there would correspond a real number - and vice versa. So we label the sequences under this supposed bijection;

$s^{r_1} , s^{r_2} , ... $

From the subset A we pick the elements (distinct) a and y.

Now consider the sequence with first value x if $s^{r_1}_{r_1}$ is y y if x, second value y if $s^{r_2}_{r_2}$ is x and vice versa, and "j'th value" x if $s^{r_j}_{r_j}$ is y and vice cersa . Then this sequence is not on the list so there was no bijection after all.

That look okay guys?

Adam
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    Are you asking whether the set of well-orderings of the real numbers has cardinality strictly larger than the set of real numbers? – Arturo Magidin Feb 02 '12 at 20:54
  • You are running into a potentially confusing issue: your well-ordering is given as a list, and so should be indexed by an ordinal; and that is more or less how you describe it; but you are saying that this "list" is "a function from the reals to the reals"; the problem is that if you are viewing it as a function from the reals to the reals, then there is no "first" value, no "second value", no "third value", etc., so that talking about "the start" does not make sense. You can fix this by first selecting a well-ordering for $\mathbb{R}$ and using that as a basis. – Arturo Magidin Feb 02 '12 at 20:59
  • Thanks for the reply Arturo - actually yes I would be interested in that question also, however for now I want to see if the (edited) version of the above has applied the diagonal argument correctly. For what I see, if we take a given set X and fix a well order (for X), we can use Cantor's diagonal argument to specify if a certain type of set (such as the function with domain X and image of {0,1} has the same cardinilty as X. – Adam Feb 02 '12 at 21:07
  • Cantor's Theorem is the statement that if $X$ is any set, and $f\colon X\to P(X)$ is any function from $X$ to its power set, then $f$ is not onto; the proof is to consider the set ${y\in X\mid y\notin f(y)}$ and show it is not in the image of $f$. This is the general version of "Cantor's diagonal argument", but it doesn't need you to consider well orderings (the existence of well orderings requires the Axiom of Choice, whereas Cantor's Theorem holds even in the absence of the Axiom of Choice). (cont) – Arturo Magidin Feb 02 '12 at 21:13
  • I edited again to be hopefully clear - if anything needs more elaboration please say so, I find it hard to describe exactly the thing I want to be shown. – Adam Feb 02 '12 at 21:13
  • (cont) I can't really make out whether you are using the Diagonal Argument "correctly", because I don't understand what it is you are trying to show. Hence my question trying to figure out what your question is. Your reply seems to indicate that I did not interpret your intention correctly. – Arturo Magidin Feb 02 '12 at 21:14
  • I'm not sure what you are trying to prove. Could you please state this clearly? – Asaf Karagila Feb 02 '12 at 21:15
  • Let X be a set and A a set with at least two elements. Then the set of all functions from X to A does not have the same cardinilty as X (call the set S). I know this is true as we can find a bijection from (a subset of) S to P(X) (because {0,1}^X has the same cardinilty as P(X) for example). I wanted to use some kind of diagonal argument for the proof however, though from what I could gather it only works if you can somehow order the set X, which allows you to choose a new element of A at every point of X for this new function that is not in the bijectiion. – Adam Feb 02 '12 at 21:21
  • in my question I was trying to be concrete and use an example (the first one I thought up before coming to the idea) as I assumed it would be easier - I am not very good with explanations though so the confusion is my fault. – Adam Feb 02 '12 at 21:23
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    @Adam: The definition ${y\in X\mid y\notin f(y)}$ is a "diagonal argument". Intuitively, you "list" the elements of $X$ on the vertical of a "table", and you "list" the subsets of $X$ on the "horizontal" of a "table" (your "well-ordering") and then you put a $0$ on the $(x,S)$ entry if $f(x)\notin S$ and a $1$ if $f(x)\in S$. We are constructing the set ${y\in X\mid y\notin f(y)}$ by looking at the "diagonal" entries $(y,f(y))$ and "switching the value" to get a new subset. – Arturo Magidin Feb 02 '12 at 21:25
  • I'm unsure if I should edit the original question - do you think that would be best? – Adam Feb 02 '12 at 21:26
  • Ah right, makes more sense - thanks Arturo! – Adam Feb 02 '12 at 21:27
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    @Adam: To read more about the diagonal argument in Cantor's proof: http://math.stackexchange.com/questions/85024/namesake-of-cantors-diagonal-argument/85066#85066 – Asaf Karagila Feb 02 '12 at 21:30

2 Answers2

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Suppose that the real numbers have cardinality $\kappa$, then there are $\kappa^+$ many ordinals which are in bijection with the real numbers, and $2^{2^{\aleph_0}}$ many ways to order them in every given order type.

This means that there are a lot more ways to order the real numbers, even if you consider them up to isomorphism.

Asaf Karagila
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Let $A$ be a set with at least two elements. You may be trying to give a diagonalization proof of the fact that there is no bijection from $X$ to the set $A^X$ of all functions from $X$ to $A$.

We show there is no mapping of $X$ onto $A^X$. The argument is standard diagonalization. No ordering is used.

Let $a$ and $b$ be two of the elements of $A$.
Suppose that $\varphi$ is a function from $X$ to $A^X$. For notational convenience, write $\varphi_x$ intead of $\varphi(x)$. Define a function $f$ from $X$ to $A$ as follows.

If $\varphi_x(x)=a$, let $f(x)=b$. Otherwise, let $f(x)=a$. It is easy to verify that $f$ cannot be $\varphi_x$ for any $x\in X$. For if $f=\varphi_e$, what is $f(e)$? If $f(e)=a$, then $f(e)=b$. If $f(e)\ne a$, then $f(e)=a$.

So given any mapping $\varphi$ of $X$ into $A^X$, we have produced $f\in A^X$ (where of course $f$ depends on $\varphi$) such that $f$ is not in the image of $\varphi$. So $\varphi$ cannot be onto, and therefore cannot be a bijection.

André Nicolas
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