# How to enumerate points in an interval as part of the proof of the Extreme Value Theorem?

The proof of Extreme Value Theorem (part 1) states:

If $$f(x)$$ is continuous on $$[a,b]$$, then it is bounded on $$[a,b]$$.

I'm confused with this part of the proof:

Suppose $$f(x)$$ is not bounded; then, for every natural number $$n$$, there exists an $$x_n \in [a,b]$$, such that $$f(x_n) > n$$

My question is, how we can give some order to rational numbers in this interval $$[a,b]$$? We cannot count rational numbers.

Is this enumeration done with respect to the magnitude of numbers from $$[a,b]$$, or is it just random enumeration?

Can we just start the proof like this?

Suppose $$f(x)$$ is not bounded; then there is no natural number that is greater than or equal to $$f(x)$$ where $$x \in [a,b]$$?

• Where do you see that the proof is talking about rational numbers ?? – TheSilverDoe Aug 30 '20 at 14:34
• Well, you do not need to use the rationals for the theorem, but you can enumerate the rationals by finding an injection $\mathbb Q\mapsto \mathbb N.$ One such is $(p,q)\mapsto 2^p3^q.$ – Matematleta Aug 30 '20 at 14:46
• The numbers in $[a,b]$ need not be rational: you are enumerating them via the natural numbers, which do have an ordering. – Integrand Aug 30 '20 at 14:49
• The numbers $\{x_n\}$ do not enumerate all numbers in $[a,b]$. The proof is choosing a very specific (countable) sequence of numbers from $[a,b]$. Specifically, given a natural number $n$, you get to choose some $x\in [a,b]$ such that $f(x)>n$. This $x$ exists by the assumption that $f$ is unbounded on $[a,b]$. But different choices of $n$ could lead to different $x$'s. So we label them by $x_n$, and the result is a sequence $(x_n)_{n=1}^{\infty}$ from $[a,b]$. – halrankard2 Aug 30 '20 at 14:53
• What numbers we a dealing with then? What are the things that [a,b] contains? – nutella_eater Aug 30 '20 at 22:23

Make a graph of an unbounded function on $$[a,b]$$. Draw horizontal lines $$y=n+1/2$$ for each integer. Since $$f$$ is unbounded on $$[a,b],$$ each of these lines must intersect the graph of $$f$$ at least once on $$[a,b]$$, so for each integer $$n$$ we get an output $$y_n$$ of $$f$$, which in turn gives us an input $$x_n\in [a,b]$$ such that $$y_n=f(x_n)=n+1/2$$.
In this way, we get, for each integer $$n$$, a number $$x_n\in [a,b]$$ such that $$f(x_n)>n$$. That is, we have associated with each integer $$n$$, a value of $$f$$ which satisfies the given condition.