# Starting Bisection Proof of Extreme Value Theorem

I am having difficulties beginning a proof for the following statement:

Use a proof strategy of bisection to prove that every function $f:[a,b] \to \mathbb{R}$ that is not bounded above is discontinuous at some point $c \in [a,b]$ (and discontinuous from the right or left if $c=a$ or $b$, respectively.

Although the strategy is given, I am having trouble getting started. Furthermore, how would I use this bisection argument to prove that if $f:[a,b] \to \mathbb{R}$ is continuous and $\sup{\{f(x):a \leq x \leq b\}}=M$, then $f(c)=M$ for some $c \in [a,b]$. I believe that this is a reformulation of the Extreme Value Theorem.

Many thanks in advance. I am using the textbook Introduction to Analysis by Arthur Mattuck.

If $f:[a,b] \to \mathbb{R}$ is not bounded above, there is a point $x_1 \in [a, b]$ such that $f(x_1) > 1$.
Divide $[a, b]$ into two parts $[a, x_1]$ and $[x_1, b]$. In one of these parts, there is an $x_2$ such that $f(x_2) > 2$ (by the unboundedness of $f$). The width of this interval is $\le \frac{b-a}{2}$.
Again, divide that part at $x_2$ into two parts. The width of each of these parts is $\le \frac{b-a}{4}$, and there must be an $x_3$ in one of these parts such that $f(x_3) > 3$ (or $f(x_3) >$ some large value if you want).
Repeating this, after the $n$-th division, there is an interval or width $\le \frac{b-a}{2^{n-1}}$ with a point $x_n$ such that $f(x_n) > n$. This sequence of points $(x_n)$ converges to a limit and $f(x_n)$ is unbounded, so $f$ can not be continuous at the limit.
• My only question is was $f(x_1)>1$ chosen arbitrarily? That is, was the lower bound 1 selected since it is the first member of $\mathbb{N}$? – Jamil_V Oct 30 '13 at 2:40