Monotonic

Monotonically Increasing

A sequence of real numbers $\{p_n\}$ is called monotonically increasing if $\forall n\in \mathbb{N},p_n\le p_{n+1}$.

• $p_1\le p_2\le p_3\le \cdots$

Monotonically Decreasing

A sequence of real numbers $\{p_n\}$ is called monotonically decreasing if $\forall n\in \mathbb{N},p_n\ge p_{n+1}$.

• $p_1\ge p_2\ge p_3\ge \cdots$

Lemma (Monotone Convergence Theorem for Sequences)

• Suppose $\{p_n\}$ is monotonically increasing and bounded from above. Then $\{p_n\}$ converges.
• If $\{p_n\}$ is monotonically decreasing and bounded from below, then $\{p_n\}$ converges.

Proof:

We only need to prove the first part.

P1 = P2    P3 = P4    P5    P6  P7      P8         l        M
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Let $\epsilon >0$. Let $A=\{p_n\mid n\in \mathbb{N}\}$. Then $A\subset \mathbb{R},A\ne \varnothing$ since $p\in A$. We know it is bounded above. Thus, it has a Supremum, $\ell :=\sup (A)$. I claim that

$$p_n\to \ell$$

Consider $\ell -\epsilon <\ell$. Then $\ell -\epsilon$ is not a upper bound for $A$. Thus, $\exists N\in \mathbb{N}$ such that $\ell -\epsilon <p_N$. On the other hand, since $\{p_n\}$ is monotonically increasing,

$$\forall n\ge N,\ell -\epsilon \le p_N\le p_n\le \ell$$

so $\forall n\ge N$,

$$\begin{array}{rl}\ell -\epsilon <p_n\le \ell & ⟹-\epsilon <p_n-\ell \le 0<\epsilon \\ & ⟹\forall n\ge N,|p_n-\ell |<\epsilon \end{array}$$

And so it converges.

Remark (Monotonic Space)

When a sequence is monotonic, then

$$\text{Convergent}⟺\text{Bounded}$$

Definition (Monotonic Functions)

Let $f:(a,b)\to \mathbb{R}$. We say $f$ is monotonically increasing if

$$\forall x,y\in (a,b),(x<y)⟹f(x)\le f(y)$$

Similarly, $f$ is monotonically decreasing if

$$\forall x,y\in (a,b),x<y⟹f(x)\ge f(y)$$

It is strict if we replace $\le$ with $<$ and $\ge$ with $>$.

Theorem (Monotonic Gives Limits)

Let $f$ be a monotonically increasing function on $(a,b)$. Then $\forall x\in (a,b)$ we have

1. $f(x-)$ and $f(x+)$ exist.
2. $f(x-)\le f(x)\le f(x+)$. Moreover, $x<y⟹f(x+)\le f(y-)$.

Proof: We show that $f(x+)$ exists and satisfied $f(x)\le f(x+)$. Let $A=\{f(t)\mid x<t<b\}$. We note that $A$ is nonempty. So, $\forall t\in (x,b)$ since $f$ is monotonically increasing, $f(x)\le f(t)$, so it is bounded below. Thus, $\inf A=\ell$ exists by Completeness Axiom.

Claim $\ell =f(x+)$. Let $\epsilon >0$.

• For $t\in (a,b)$, we need to find $\delta >0$ such that if $x<t<x+\delta$, then $|f(t)-\ell |<\epsilon$.
• Since $\epsilon >0$ and $\ell =\inf A$, then $\ell +\epsilon >\ell$ and so it is not a lower bound for $A$.
• Thus $\exists x<t_0<b$ such that $\ell \le f(t_0)<\ell +\epsilon$ since $\ell +\epsilon$ not LB, and density. Now for all $t$ such that $x<t<t_0<b$, we have

1. $f(t)\in A⟹\ell \le f(t)$
2. $f(t)\le f(t_0)$ and $f$ monotonically increasing, so $f(t)<\ell +\epsilon$ Thus, $\forall x<t<t_0$ we have $|f(t)-\ell |<\epsilon$.

So if $\delta =t_0-x$ then we are done. A similar argument shows for $f(x-)\le f(x)$. Now suppose

$$a<x<y<b$$

Then

$$f(x+)=\inf \{f(t)\mid x<t<b\}=\inf \{f(t)\mid x<t<y\}$$

and

$$f(y-)=\sup \{f(s)\mid a<s<y\}=\sup \{f(s)\mid x<s<y\}$$

Then

$$\begin{array}{rl}f(x+)& =\inf \{f(t)\mid x<t<y\}\\ & \le \sup \{f(s)\mid x<s<y\}\\ & =f(y-)\end{array}$$

And so they are equal.

Corollary (Discontinuity by Monotonic)

If $f$ is monotonic, then all discontinuities are of the first kind.

Proposition (Countable Discontinuities)

If $f$ is monotonic, then the set of discontinuities is at most countable.

Proof: Let us assume $f$ is monotone. By theorem,

$$(\forall x\in (a,b))(f(x-)\le f(x)\le f(x+))$$

So if $f$ is discontinuous at $x$, then

$$f(x-)<f(x+)⟺(f(x-),f(x+))\text{ is a nonempty open interval}$$

Now let $x,y\in (a,b)$ with $x<y$. If $f$ is discontinuous at $x,y$. Then by theorem, $f(x+)\le f(y-)$. So we get:

$$(1)(f(x-),f(x+))\cap (f(y-),f(y+))=\varnothing$$

Let $A=\{x\in (a,b)\mid f\text{ discontinuous at }x\}$. If $A$ is nonempty, then $\forall x\in A$, let

$$r(x)\in (f(x-),f(x+))\cap \mathbb{Q}$$

which exists be density of $\mathbb{Q}$. Define

$$\phi :A\to \mathbb{Q}x\mapsto r(x)$$

We show that $\phi$ is injective. Note for $x<y⟹r(x)<r(y)$ by $(1)$. Since $\mathbb{Q}$ is countable, $\phi$ injective, then $A$ is at most countable.

Proposition

Given any at most countable subset $A=(a_n{)}_{n\to \mathbb{N}}\subset (a,b)$, there exists an increasing function on $(a,b)$ such that the discontinuity of $f$ are exactly $A$.