Continuous function/R/Intermediate value theorem/Fact/Proof

Proof

We consider the situation ${}f(a)\leq u\leq f(b)$, and show the existence of such an ${}x$ with the bisection method. For that, we put ${}a_{0}:=a$ and ${}b_{0}:=b$, we consider the arithmetic mean ${}x_{0}:={\frac {a_{0}+b_{0}}{2}}$, and we compute

$f{\left(x_{0}\right)}.$

If ${}f{\left(x_{0}\right)}\leq u$, then we put

$a_{1}:=x_{0}\,\,{\text{ and }}\,\,b_{1}:=b_{0}$

and if ${}f{\left(x_{0}\right)}>u$, then we put

$a_{1}:=a_{0}\,\,{\text{ and }}\,\,b_{1}:=x_{0}.$

In each case, the new interval ${}[a_{1},b_{1}]$ is lying inside the initial interval and has half of its length. It fulfills again the condition ${}f{\left(a_{1}\right)}\leq u\leq f{\left(b_{1}\right)}$, therefore we can apply the same defining method again and get recursively a family of nested intervals. Let ${}x$ denote the real number which is determined by these nested intervals. For the lower bounds of the intervals, we have ${}f{\left(a_{n}\right)}\leq u$, and this carries over to the limit ${}x$, due to the criterion for continuity in terms of sequences. Hence, ${}f{\left(x\right)}\leq u$. For the upper bounds, we have ${}f{\left(b_{n}\right)}\geq u$, and this again carries over to ${}x$, so ${}f(x)\geq u$.  Therefore, ${}f(x)=u$.

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