Proof of necessary condition of injection without strict monotonicity

Question

According to the proof details in Continuous Injection of Interval is Strictly Monotone, I deduce a statement:

Statement: if an injective function $f \colon D\subset \mathbb{R}\to \mathbb{R}$ is not strictly monotone, then there exist $x,y,z∈D$ with $ x<y<z$ such that either $f(x)<f(y)$ and $f(y)>f(z),$ or $f(x)>f(y)$ and $f(y)<f(z).$

But I have trouble in proving this statement, or even disprove it. My trial is following.

Suppose otherwise the conclusion of the statement is not true, then we have \begin{align*} \forall x, y, z\in D: x<y<z \implies (f(x)>f(y) \lor f(y)<f(z)) \land (f(x)<f(y)\lor f(y)>f(z)). \end{align*} Because \begin{align*} &(f(x)>f(y) \lor f(y)<f(z)) \land (f(x)<f(y)\lor f(y)>f(z))\\ \iff&(f(x)>f(y)\land f(x)<f(y))\lor (f(x)>f(y)\land f(y)>f(z))\lor (f(y)<f(z)\land f(x)<f(y))\lor (f(y)<f(z)\land f(y)>f(z))\\ \iff &(f(x)>f(z))\lor (f(z)>f(x)), \end{align*} and for all statements $P, Q, R,$
\begin{gather*} P\implies (Q\lor R) \text{ is logically equivalent to } (P\implies Q) \lor (P\implies R), \end{gather*} we have $$\forall x, y, z\in D: x<y<z \implies (f(x)>f(y) \lor f(y)<f(z)) \land (f(x)<f(y)\lor f(y)>f(z))$$ is logically equivalent to \begin{gather*} \forall x,z\in D: (x<z\implies f(x)>f(z)) \lor (x<z\implies f(x)<f(z)). \end{gather*} But we know that the statement $$\forall x,z\in D: (x<z\implies f(x)>f(z)) \lor (x<z\implies f(x)<f(z))$$ is not equivalent to \begin{gather*}\tag{$\star$}(\forall x,z\in D: x<z\implies f(x)>f(z)) \lor (\forall x,z\in D: x<z\implies f(x)<f(z))!\end{gather*} As you can see, $(\star)$ is the definition that $f$ is strictly increasing, or strictly decreasing on $D.$ Thus I was stuck!

But if the domain $D$ of definition of $f$ is an interval, and $f$ is continuous on $D,$ I have proved that the statement is true. Thus I guess that maybe the continuity and connectedness conditions are redundant for this statement.

So, if you think that the statement is true, how to prove it. If it is false, can you construct a counterexample?

Answer 1

Since $f$ is not strictly decreasing, there are some $a, b \in D$, such that $a < b$ and $f(a) \leq f(b)$. Since $f$ is injective, $f(a) < f(b)$. Since $f$ is not strictly increasing, there are some $c, d \in D$, such that $c < d$ and $f(c) \geq f(d)$. Since $f$ is injective, $f(c) > f(d)$.

Case 1: $a < b < c < d$. If $f(b) < f(c)$, set $(x,y,z) := (b,c,d)$. Otherwise, set $(x,y,z) := (a,b,c)$.

Case 2: $a < b = c < d$. Set $(x,y,z) := (a,b,c)$.

Case 3: $a < c < b < d$. If $f(a) < f(c)$, set $(x,y,z):=(a,c,d)$. Otherwise, set $(x,y,z):=(a,c,b)$.

Case 4: $a < c < b = d$. Set $(x,y,z):=(a,c,b)$.

Case 5: $a < c < d < b$. If $f(a) < f(c)$, set $(x,y,z):=(a,c,d)$. Otherwise, set $(x,y,z):=(a,c,b)$.

Case 6: $a = c < b < d$. Set $(x,y,z):=(a,b,d)$.

Case 7: $a=c < b = d$. Impossible!

Case 8: $a=c < d < b$. Set $(x,y,z):=(a,d,b)$.

Case 9: $c < a < b < d$. If $f(c) < f(a)$, set $(x,y,z):=(c,a,d)$. Otherwise, set $(x,y,z):=(c,a,b)$.

Case 10: $c < a < b = d$. Set $(x,y,z):=(c,a,b)$.

Case 11: $c < a < d < b$. If $f(c) < f(a)$, set $(x,y,z):=(c,a,d)$. Otherwise, set $(x,y,z):=(c,a,b)$.

Case 12: $c < a = d < b$. Set $(x,y,z):=(c,a,b)$.

Case 13: $c < d < a < b$. If $f(d) < f(a)$, set $(x,y,z):=(c,d,a)$. Otherwise, set $(x,y,z):=(d,a,b)$.

Answer 2

Here's a simple proof that avoids all those mind-numbing cases!

Lemma: Let $D\subseteq \mathbb{R}.$ If $g\colon D\to\mathbb{R}$ is strictly monotonic on every 3-element subset of $D,$ then $g$ is strictly monotonic on every subset of $D$ of cardinality $\le4.$

Proof of Lemma: It's sufficient to prove the conclusion for subsets of cardinality exactly $4,$ since it's trivially true for sets of size $3$ or less. So let $a \lt b \lt c\lt d$ be elements of $D.$ Then $g$ is strictly monotonic on both $\lbrace a, b, c \rbrace$ and $\lbrace b, c, d \rbrace.$ In fact, if $g(b)\lt g(c),$ then $g$ is strictly increasing on both those $3-$element sets; and if $g(b)\gt g(c),$ then $g$ is strictly decreasing on both those $3-$element sets. It follows that $g$ is either strictly increasing or strictly decreasing on the $4$-element set $\lbrace a, b, c,d \rbrace.$ $$ $$ The problem now follows immediately: Assume $f$ is injective and not strictly monotonic. Then:

$\bullet$ There exist $a$ and $b$ such that $a\lt b$ and $f(a) \gt f(b)$ (since $f$ isn't strictly increasing).

$\bullet$ There exist $c$ and $d$ such that $c\lt d$ and $f(c) \lt f(d)$ (since $f$ isn't strictly decreasing).

But now $\lbrace a, b, c, d \rbrace$ is a subset of $D$ of cardinality at most $4$ on which $f$ isn't strictly monotonic. So, applying the lemma, there's some $3$-element subset of $D$ on which $f$ is not strictly monotonic, which is exactly what we need to prove.