Continuum Property implies Well-Ordering Principle

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The Continuum Property of the positive real numbers $\R_{\ge 0}$ implies the Well-Ordering Principle of the natural numbers $\N$.


Suppose that the set of positive real numbers $\R_{\ge 0}$ has the Continuum Property.

Aiming for a contradiction, suppose $T \subseteq \N$ is a subset of $\N$ which has no smallest element.


$\forall t \in T: \exists u \in T: u < t$

Let $A \in \R_{\ge 0}$.

For every $t \in T$, let $a_t = \dfrac A {2^t}$.

We have that $t \in \N$, so $t \ge 0$.

Therefore $2^t \ge 1$.

This $a_t$ can never be greater than $A$.

Therefore $\set {a_t: t \in T}$ has an upper bound.

As $\R_{\ge 0}$ has the Continuum Property, it follows that $\set {a_t: t \in T}$ has a supremum $B$.

We have that $\dfrac {7 B} 8 < B$.

Thus $\dfrac {7 B} 8$ is not an upper bound of $\set {a_t: t \in T}$.

Therefore there exists $m \in T$ such that:

$a_m > \dfrac {7 B} 8$

Then there is $n \in T$ such that $n < m$.

Let $n = m - p$.


\(\ds a_n\) \(=\) \(\ds \dfrac A {2^n}\)
\(\ds \) \(=\) \(\ds \dfrac A {2^{m - p} }\)
\(\ds \) \(=\) \(\ds \dfrac {A \cdot 2^p} {2^m}\)
\(\ds \) \(\ge\) \(\ds \dfrac {A \cdot 2} {2^m}\)
\(\ds \) \(=\) \(\ds 2 \cdot a_m\)
\(\ds \) \(>\) \(\ds 2 \cdot \dfrac {7B} 8\)
\(\ds \) \(=\) \(\ds \dfrac {7B} 4\)
\(\ds \) \(>\) \(\ds B\)

Therefore $a_n > B$.

This contradicts the supposition that $B$ is the supremum of $\set {a_t: t \in T}$

Thus our initial supposition that $T \subseteq \N$ has no smallest element was false.

As $T$ is arbitrary, it follows that each $T \subseteq \N$ has a smallest element.

Thus the Well-Ordering Principle holds for $\N$, as was to be proved.


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