Exercise 2.6.7

Question

Exercises 2.4.4 and 2.5.4 establish the equivalence of the Axiom of Completeness and the Monotone Convergence Theorem. They also show the Nested Interval Property is equivalent to these other two in the presence of the Archimedean Property.
  \begin{enumerate}[label=(\alph*)] 
  \item Assume the Bolzano-Weierstrass Theorem is true and use it to construct a proof of the Monotone Convergence Theorem without making any appeal to the Archimedean Property. This shows that BW, AoC, and MCT are all equivalent.
  \item Use the Cauchy Criterion to prove the Bolzano-Weierstrass Theorem, and find the point in the argument where the Archimedean Property is implicitly required. This establishes the final link in the equivalence of the five characterizations of completeness discussed at the end of Section $2.6$.
  \item How do we know it is impossible to prove the Axiom of Completeness starting from the Archimedean Property?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Suppose $(x_n)$ is increasing and bounded, BW tells us there exists a convergent subsequence $(x_{n_k}) 	o x$, We will show $(x_n) 	o x$. First note $x_k \le x_{n_k}$ implies $x_n \le x$ by the Order Limit Theorem.

Pick $K$ such that for $k \ge K$ we have $|x_{n_k} - x| < \epsilon$.
    Since $(x_n)$ is increasing and $x_n \le x$ every $n \ge n_K$ satisfies $|x_n - x| < \epsilon$ as well.
    Thus $(x_n)$ converges, completing the proof.
  \item We're basically going to use the Cauchy criterion as a replacement for NIP in the proof of BW.
    Recall we had $I_{n+1} \subseteq I_n$ with $a_{n_k} \in I_k$, we will show $a_{n_k}$ is Cauchy.

The length of $I_k$ is $M(1/2)^{k-1}$ by construction, so clearly $|a_{n_k} - a_{n_j}| < M(1/2)^{N-1}$ for $k,j \ge N$, implying $(a_{n_k})$ converges by the Cauchy criterion.

We needed the Archimedean Property to conclude $M(1/2)^{N-1} \in \mathbf{Q}$ can be made smaller then any $\epsilon \in \mathbf{R}^+$.
  \item The Archimedean Property is true for $\mathbf{Q}$ meaning it cannot prove AoC which is only true for $\mathbf{R}$. (If we did, then we would have proved AoC for $\mathbf{Q}$ which is obviously false.)
   \end{enumerate}

Exercise 2.6.7

Answers

Comments

Exercise 2.6.7

Answers

Comments

Add answer