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Let $f: [0,1]\to [0,1]$ be continuous and one-to-one. Prove that $f$ is either strictly increasing or strictly decreasing. Sorry if this is a duplicate question. Not sure whether or not to prove this by contradicting the fact that $f$ is one-to-one.

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  • $\begingroup$ If you assume that $f$ is one-to-one and monotone, you'll have an easy proof by contradiction. $\endgroup$ – Demosthene Mar 25 '15 at 18:00
  • $\begingroup$ And for a direct proof, use the definition of injectivity of a function. See my answer for explanation. $\endgroup$ – Prasun Biswas Mar 25 '15 at 18:01
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Think of it like this.

Since $f$ is one-one, w.l.o.g., assume $x_1\lt x_2$ be two elements of $\textrm{Dom}(f)$.

Then, $x_1\neq x_2\implies f(x_1)\neq f(x_2)$.

By the trichotomy principle, either $f(x_1)\lt f(x_2)$ or $f(x_1)\gt f(x_2)$.

This is sufficient to prove that $f$ is either strictly increasing or strictly decreasing.

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Let $x_1, x_2 \in [0,1]$ (in your domain). The primary definition of $f$ being one-to-one is as follows: If $x_1 \not= x_2$, then $f(x_1)\not=f(x_2)$. So either $x_1 < x_2$ implies $f(x_1)<f(x_2)$ (ao $f$ is strictly increasing) or $x_1 < x_2$ implies $f(x_1) > f(x_2)$ (so $f$ is strictly decreasing). Note that this conclusion follows only if $f$ is continuous on your domain.

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Suppose that e.g. $a<b<c$ and $f(a)<f(b)>f(c)$.

According to the intermediate value theorem for any $y$ that satisfies $\max(f(a),f(c))<y<f(b)$ there are values $u\in(a,b)$ and $v\in(b,c)$ such that $f(u)=y=f(v)$. This contradicts the injectivity of $f$.

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Define for $x\in[0,1]$, $h\in(0,1-x]$, $$F(x,h)=f(x+h)-f(x)$$ Since $F$ is continuous, its domain is connected and does not vanish, it must be positive or negative.

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