Reading Notes-Some basic theorems for real numbers

Axiom of completeness

Any nonempty set of real numbers that is bounded above has an least upper bound.

Monotone Convergence Theorem

Let $(a_n)$ be an increasing sequence that is bounded above, then it converges.

Proof:

Consider the set of point $\{ a_n : n \in N \}$. By assumption, it is bounded. By Axiom of Completeness, the set $\{ a_n \}$ has an least upper bound, denoted as $a = \sup \{ a_n : n \in N \}$. Because $a$ is a least upper bound, $\forall \epsilon > 0$, there is an element $a_N$, such that $a_N \ge a - \epsilon$. But $(a_n)$ is an increasing sequence. Therefore, for any $n \ge N$, we have $a \ge a_n \ge a_N \ge a - \epsilon$, which implies that $(a_n) \rightarrow a$.

Corollary:

Let $(b_n)$ be a decreasing sequence that is bounded below, then it converges.

Nested Interval Property

Let $I_1 \supset I_2 \supset I_3 ...$ be a collection of closed intervals. Then $\cap_{n} I_n \neq \emptyset$.

Proof 1 (By Axiom of Completeness):

Denote $I_n = [a_n, b_n]= \{ x \in R : a_n \le x \le b_n \}$. By definition, we have

We are going to use Axiom of Completeness to produce a single real number $a \in \cap_n I_n$. The set we consider is

It is bounded above by any $b_n$ for $n \ge 1$. Therefore we are justified in the setting $a = \sup A$. Because $a$ is an upper bound, we have $a_n \le a$. The fact $b_n$ is an upper bound and that $a$ is the least upper bound implies $a \le b_n$. Putting together, $a \in I_n$ for $n \ge 1$. Hence, $a \in \cap_n I_n$.

Proof 2 (By Monotone Convergence Theorem):

$(a_n)$ is a monotone increasing sequence and bounded above. By Monotone Convergence Theorem, there exists a real number $a = \lim_n a_n$. Moreover, it holds that $b_m \ge a_n$ for any integer $m$ and $n$, which implies that $b_m \ge \lim_n a_n = a$, as desired.

Remark:

We may not necessarily choose Axiom of Completeness as our starting point. For example, if we assume that Nested Interval Property is true and $\lim 1/2^n \rightarrow 0$, it can be used to prove Axiom of Completeness.

Proof: Let $A$ be an nonempty set that is bounded above, let $b_1$ be an upper bound and $a_1$ be an element in $A$. Necessarily, $a_1 < b_1$ and we can define the closed interval $I_1 = [a_1, b_1]$.

The idea is to construct a sequence of closed intervals and use Nested Interval Property. Now we bisect $I_1$ into two closed intervals of equal length let $c_1$ be the midpoint. If $c_1$ is an upper bound for $A$, we define $I_2 = [a_1, c_1]$. Otherwise, let $I_2 = [c_1, b_1]$. In general, we construct $I_n = [a_n, b_n]$ by taking the half from $I_{n - 1}$ such that, $a_n$ is no greater than any upper bound of $A$ and $b_n$ is no less than any element of $A$.

By Nested Interval Property, there exists $a \in \cap_n I_n$. Also we notice that the length of $I_n \rightarrow 0$, which implies $\lim a_n = \lim b_n = s$.

Because for all $n \in N$, and any element $a \in A$, we have $a \le b_n$. Taking limit of the right hand side, we obtain $a \le s$, which shows $s$ is an upper bound for $A$.

Moreover, for any upper bound $l$ for $A$, we have $a_n \le l$. Again taking limit of the left hand side, we conclude $s \le l$, which shows that $s$ is the least upper bound.

Bolzano-Weierstrass Theorem

Let $(a_n)$ be a bounded sequence, then it has a convergent sub-sequence.

Proof:
Since $(a_n)$ is bounded there exists $M > 0$ such that $|a_n| \le M$ for all $n \in N$ . Define $I_0 = [-M, M]$. Then at least one of the intervals $[-M, 0]$ and $[0, M]$ contains infinite number of terms from the sequence $(a_n)$. Select such half interval and label it $I_1$. Let $a_{n_1}$ be some term from $(a_n)$ belonging to $I_1$.

Next, we continue to bisect $I_1$ into two closed intervals of equal length. Choose the half that contains infinite number of terms from $(a_n)$ and label it as $I_2$. We can select an element $a_{n_2} \in I_2$ from $(a_n)$ with $n_2 \ge n_1$. In general, we construct $I_k$ by taking the half from $I_{k - 1}$ that contains infinite number of terms of $(a_n)$ and then choose $a_{n_k} \in I_k$ so that $n_k > n_{k - 1} > ... > n_2 > n_1$. The sets $\{ I_n : n \in N \}$ forms a sequence of nested closed intervals

By Nested Interval Property, there is at least on point $a \in \cap_n I_n$. On the other hand, the lengths of the intervals shrink to $0$, which implies that $a_{n_k} \rightarrow a$, as desired.

Cauchy Criterion

Let $(a_n)$ be a sequence which satisfies

Then $(a_n)$ converges.

Proof:

Step 1: $(a_n)$ is bounded. To see this, let $N_1$ be the integer that for $n \ge N$, we have $|a_n - a_{N_1} | \le 1$ and so $|a_n| \le |a_{N_1} | + 1$. It can be verified $\max \{ |a_1|, |a_2|, ..., |a_{N_1 - 1} |, |a_{N_1} | + 1 \}$ is an upper bound for the sequence $( a_n )$.

Step 2: Now, by Bolzano-Weierstrass Theorem, $(a_n)$ has an convergent sub-sequence $( a_{n_k} ) \rightarrow a$. We show that $(a_n)$ converges to the same limit. Let $\epsilon > 0$, because $(a_n)$ is Cauchy,

We also know $(a_{n_k}) \rightarrow a$, so choose a term $a_{n_k}$, with $n_k > N$ and

Putting together,

as desired.

Remark:

Cauchy Criterion can be used to prove Bolzano-Weierstrass Theorem. The key is to construct a sequence of nested closed intervals $I_1 \supset I_2 \supset I_3 ...$ as before and obtain a sub-sequence $(a_{n_k})$. Then, instead of using Nested Interval Property, we observe that $(a_{n_k})$ is Cauchy as the length of $I_k$ shrinks to zero. We know $(a_{n_k})$ converges by Cauchy Criterion.

Reference: Stephen Abbott, Understanding Analysis