Theorem 5

Let I be an interval in R. Then a mapping f of I into R is a homeomorphism of I onto f(I) if and only if f is strictly monotonic and continuous on I; and f(I) is then an interval in R.

1) The condition is necessary. Let $a$ and $b$ be two points of I such that $a < b$, and suppose for example that $f(a) < f(b)$. Let us show that $f$ is strictly increasing on I. First, if $a < c < b$ then we must have $f(a) < f(c) < f(b)$; if, for example, we had $f(a) < f(b) < f(c)$ then the image of the interval $[a, c]$ under $f$ would be a connected set (Chapter I, , no. 2, Proposition 4) and would therefore contain the interval $[f(a), f(c)]$; hence there would exist $x \in [a, c]$ such that $f(x) = f(b)$, contrary to the hypothesis that $f$ is injective

It follows that if $x$ and $y$ are any two points of I such that $x < y$, then $f(x) < f(y)$; for we have $f(a) < f(x) < f(b)$ if $a < x < b$, $f(a) < f(b) < f(x)$ if $b < x$, and $f(x) < f(a) < f(b)$ if $x < a$; repeating the argument with $a, x, y$ in place of $a, b, x$ respectively, we see that $f(x) < f(y)$.

2) The condition is sufficient. Suppose that $f$ is continuous and strictly monotonic on I (say, strictly increasing): $f(I)$ is connected and is therefore an interval, and since $f$ is strictly increasing, $f$ is a bijective mapping of I onto $f(I)$. Moreover, the image under $f$ of an open interval in I is an open interval in $f(I)$, and therefore (§ 1, no, 4, Proposition 5) $f$ is a homeomorphism of I onto $f(I)$.