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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 $x \in [a,b]$?

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    $\begingroup$ Where do you see that the proof is talking about rational numbers ?? $\endgroup$ – TheSilverDoe Aug 30 '20 at 14:34
  • $\begingroup$ 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 $(p,q)\mapsto 2^p3^q.$ $\endgroup$ – Matematleta Aug 30 '20 at 14:46
  • $\begingroup$ The numbers in $[a,b]$ need not be rational: you are enumerating them via the natural numbers, which do have an ordering. $\endgroup$ – Integrand Aug 30 '20 at 14:49
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    $\begingroup$ 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 $[a,b]$. $\endgroup$ – halrankard2 Aug 30 '20 at 14:53
  • $\begingroup$ What numbers we a dealing with then? What are the things that [a,b] contains? $\endgroup$ – nutella_eater Aug 30 '20 at 22:23
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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.

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  • $\begingroup$ my problem was, that I've started from the wrong end. Your answer bring the light to the correct approach. Thanks a lot $\endgroup$ – nutella_eater Aug 30 '20 at 22:25
  • $\begingroup$ You are welcome, glad to help. $\endgroup$ – Matematleta Aug 31 '20 at 4:51

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