# Thread: Total Ordering on Real Function

1. ## Total Ordering on Real Function

I been trying to find a total ordering on $[0,1]^{[0,1]}$, the set of all functions from $[0,1]\mapsto [0,1]$. This cardinality of this set is $2^{ 2^{\aleph_0}} > 2^{\aleph_0}$. I tried a few ideas and they all failed. I happen to know a total ordering exists by the axiom of choice (this is correct, right?). But I am trying to find an explicit ordering on this set for a counterexample.

2. One of the most important consequences of the Axiom of Choice is the Well –Ordering Theorem: Every set can be well ordered. However, the proof is an outstanding example of a nonconstructive proof. That is, the proof does not give any indication of how the order relation is constructed. For example, it is not known how the set of real numbers can be well ordered.

3. Originally Posted by Plato
One of the most important consequences of the Axiom of Choice is the Well –Ordering Theorem: Every set can be well ordered. However, the proof is an outstanding example of a nonconstructive proof. That is, the proof does not give any indication of how the order relation is constructed. For example, it is not known how the set of real numbers can be well ordered.
For homework, I have the following problem:
If $(P,<)$ is a totally ordered set which has a countable dense subset then $|P|\leq 2^{\aleph_0}$
.

At first I tried proving this but failed. I assume my textbook had a mistake in it. Here is the counterexample: Let $P_1 = [0,1]^{[0,1]}$ which is totally ordered (okay, so we need the axiom of choice). Let $P_2 =\mathbb{Q}$ ordered in the natural way. Define $P = P_1\cup P_2$ where any element in $P_2$ is less than any element in $P_1$. This ordering is a total ordering and it has a countable dense subset, i.e. $P_2$, yet $|P| > 2^{\aleph_0}$. Thus, it is wrong.

But I am unsure of this counterexample. Because all the set theory we done so far never used the axiom of choice. Maybe without the axiom of choice this statement is true? Can that be? Or if it is possible to construct counterexample using axiom of choice (eventhough we never used it before) it must mean this problem is faulty. This is the reason why I tried to find an explicit ordering on the real functions, but that seems really hard to do. Do you happen to know of any total ordering for sets with cardinality greater than the countinuum?

This is Mine 93th Post!!!

4. How about the this. I been trying to find an explicit total-ordering on $[0,1]^{[0,1]}$. I found a way how to do that if we can find an explicit well-ordering of the interval $[0,1]$.

Suppose that we can explicitly well-order $[0,1]$ call it $<$. Let $f,g$ be elements of $[0,1]^{[0,1]}$. Say $f\not = g$ then the set $\Delta(f,g) = \{ x \in [0,1] : f(x) \not = g(x) \}$ is a non-empty subset of $[0,1]$ which means there is a least element $x_0$. If $f(x_0) < g(x_0)$ then we define $f \prec g$. This defines a total ordering on $\left( [0,1]^{[0,1]}, \prec \right)$. This reduces the problem to well-ordering the continuum. So are there any well-known (pun) well-orderings of the continuum?