I am reading "Calculus 4th Edition" by Michael Spivak.
The following theorem on p.234 in this book is intuitively obvious, but its proof is not so easy.
Theorem 2:
If $f$ is continuous and one-one on an interval, then $f$ is either increasing or decreasing on that interval.
Proof:
(1) If $a \lt b \lt c$ are three points in the interval, then either $f(a) \lt f(b) \lt f(c)$ or $f(a) \gt f(b) \gt f(c)$. Suppose, for example, that $f(a) \lt f(c)$. If we had $f(b) \lt f(a)$, then the Intermediate Value Theorem applied to the interval $[b,c]$ would give an $x$ with $b \lt x \lt c$ and $f(x)=f(a)$, contradicting the fact that $f$ is one-one on $[a,c]$. Similarly, $f(b) \gt f(c)$ would lead to a contradiction, so $f(a) \lt f(b) \lt f(c)$. Naturally, the same sort of argument works for the case $f(a) \gt f(c)$.(2) If $a \lt b \lt c \lt d$ are four points in the interval, then either $f(a) \lt f(b) \lt f(c) \lt f(d)$ or $f(a) \gt f(b) \gt f(c) \gt f(d)$. For we can apply (1) to $a \lt b \lt c$ and then to $b \lt c \lt d$.
(3) Take any $a \lt b$ in the interval, and suppose that $f(a) \lt f(b)$. Then $f$ is increasing: For if $c$ and $d$ are any two points, we can apply (2) to the collection $\{a,b,c,d\}$ (after arranging in increasing order).
In (3), I think Spivak assumed that $\#\{a,b,c,d\}=4$.
I don't know why.
So, I think we need to modify his proof.
So, I wrote my proof of (3) as follows.
Is my proof of (3) ok?
My proof of (3):
Take any $a < b$ in the interval.(3-1) Suppose that $f(a)<f(b).$
Let $c$ and $d$ be any two points in the interval such that $c<d$.
Obviously, $2\leq\#\{a,b,c,d\}.$(3-1-1) Suppose that $\#\{a,b,c,d\}=2.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2\}$, where $a_1<a_2.$
Since $a<b$ and $c<d$, $a=c=a_1$ and $b=d=a_2.$
So, $f(c)=f(a)<f(b)=f(d).$(3-1-2) Suppose that $\#\{a,b,c,d\}=3.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2,a_3\}$, where $a_1<a_2<a_3.$
We can write $a=a_i,b=a_j$, where $i<j.$
We can write $c=a_k,d=a_l$, where $k<l.$
By (1) in the Spivak's proof, $f(a_1)<f(a_2)<f(a_3)$ or $f(a_1)>f(a_2)>f(a_3).$
Since $f(a_i)=f(a)<f(b)=f(a_j)$ and $i<j$, $f(a_1)<f(a_2)<f(a_3)$ must hold.
Since $k<l$, $f(c)=f(a_k)<f(a_l)=f(d)$.(3-1-3) Suppose that $\#\{a,b,c,d\}=4.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2,a_3,a_4\}$, where $a_1<a_2<a_3<a_4.$
We can write $a=a_i,b=a_j$, where $i<j.$
We can write $c=a_k,d=a_l$, where $k<l.$
By (2) in the Spivak's proof, $f(a_1)<f(a_2)<f(a_3)<f(a_4)$ or $f(a_1)>f(a_2)>f(a_3)>f(a_4).$
Since $f(a_i)=f(a)<f(b)=f(a_j)$ and $i<j$, $f(a_1)<f(a_2)<f(a_3)<f(a_4)$ must hold.
Since $k<l$, $f(c)=f(a_k)<f(a_l)=f(d)$.
So, $f$ is increasing in the case (3-1).(3-2) Suppose that $f(a)>f(b).$
Let $c$ and $d$ be any two points in the interval such that $c<d$.
Obviously, $2\leq\#\{a,b,c,d\}.$(3-2-1) Suppose that $\#\{a,b,c,d\}=2.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2\}$, where $a_1<a_2.$
Since $a<b$ and $c<d$, $a=c=a_1$ and $b=d=a_2.$
So, $f(c)=f(a)>f(b)=f(d).$(3-2-2) Suppose that $\#\{a,b,c,d\}=3.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2,a_3\}$, where $a_1<a_2<a_3.$
We can write $a=a_i,b=a_j$, where $i<j.$
We can write $c=a_k,d=a_l$, where $k<l.$
By (1) in the Spivak's proof, $f(a_1)<f(a_2)<f(a_3)$ or $f(a_1)>f(a_2)>f(a_3).$
Since $f(a_i)=f(a)>f(b)=f(a_j)$ and $i<j$, $f(a_1)>f(a_2)>f(a_3)$ must hold.
Since $k<l$, $f(c)=f(a_k)>f(a_l)=f(d)$.(3-2-3) Suppose that $\#\{a,b,c,d\}=4.$
In this case we can write $\{a,b,c,d\}=\{a_1,a_2,a_3,a_4\}$, where $a_1<a_2<a_3<a_4.$
We can write $a=a_i,b=a_j$, where $i<j.$
We can write $c=a_k,d=a_l$, where $k<l.$
By (2) in the Spivak's proof, $f(a_1)<f(a_2)<f(a_3)<f(a_4)$ or $f(a_1)>f(a_2)>f(a_3)>f(a_4).$
Since $f(a_i)=f(a)>f(b)=f(a_j)$ and $i<j$, $f(a_1)>f(a_2)>f(a_3)>f(a_4)$ must hold.
Since $k<l$, $f(c)=f(a_k)>f(a_l)=f(d)$.
So, $f$ is decreasing in the case (3-2).
