proof of intermediate value theorem

We first prove the following lemma.

If $f:[a,b]\\to\\mathbb{R}$ is a continuous function with $f(a)\\leq 0\\leq f(b)$ then there exists a $c\\in[a,b]$ such that $f(c)=0$ .

Define the sequences $(a_{n})$ and $(b_{n})$ inductively, as follows.

a_{0}=a\\quad b_{0}=b

c_{n}=\\frac{a_{n}+b_{n}}{2}

(a_{n},b_{n})=\\begin{cases}(a_{n-1},c_{n-1})&f(c_{n-1})\\geq 0\\\\(c_{n-1},b_{n-1})&f(c_{n-1})<0\\end{cases}

We note that

a_{0}\\leq a_{1}\\leq\\cdots\\leq a_{n}\\leq b_{n}\\leq\\cdots\\leq b_{1}\\leq b_{0}

(b_{n}-a_{n})=2^{-n}(b_{0}-a_{0})

(1)

f(a_{n})\\leq 0\\leq f(b_{n})

(2)

By the fundamental axiom of analysis $(a_{n})\\to\\alpha$ and $(b_{n})\\to\\beta$ . But $(b_{n}-a_{n})\\to 0$ so $\\alpha=\\beta$ .By continuity of $f$

(f(a_{n}))\\to f(\\alpha)\\quad(f(b_{n}))\\to f(\\alpha)

But we have $f(\\alpha)\\leq 0$ and $f(\\alpha)\\geq 0$ so that $f(\\alpha)=0$ . Furthermore we have $a\\leq\\alpha\\leq b$ , proving the assertion.

Set $g(x)=f(x)-k$ where $f(a)\\leq k\\leq f(b)$ . $g$ satisfies the same conditions as before, so there exists a $c$ such that $f(c)=k$ . Thus proving the more general result.