Intermediate value theorem

If $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

A continuous injective real-valued function on a real interval is strictly monotone. A value outside the interval between two endpoint values would, by the intermediate value theorem, create a repeated value. Applying this observation to subintervals gives consistent strict ordering. If the left endpoint value is smaller than the right endpoint value, the function is strictly increasing; otherwise it is strictly decreasing.

If $f:[0,1]\to \mathbb{R}$ is continuous and $f(0)=f(1)$, then for every positive integer $n$ there is $\alpha \in [0,1-1/n]$ with $f(\alpha +1/n)=f(\alpha )$. For $n>1$, the continuous difference $h_n(x)=f(x+1/n)-f(x)$ satisfies ${\sum }_{j=0}^{n-1}{h}_n(j/n)=0$. Either one sampled difference is zero or two have opposite signs; the intermediate value theorem then supplies a zero between them. For $n=1$, the endpoint chord suffices.

If a continuous real-valued function on a real or complex vector space has opposite signs at two points, restricting it to the line segment between them and applying the intermediate value theorem gives a zero.

If a continuous $g:[0,na]\to \mathbb{R}$ satisfies $g(na)-g(0)=na$, then the incrementssatisfy ${\sum }_{k=0}^{n-1}h(ka)=0$. Continuity forces $h$ to vanish somewhere on $[0,(n-1)a]$.

Every continuous injective function from a real interval to the real line is strictly monotone. A continuous bijection $\mathbb{R}\to \mathbb{R}$ therefore has a continuous inverse.