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When I research on the internet, I see complicated proofs. My proof seems simple to me. Is it a valid proof?

Proof

. i understand why it was invalid. Then is it valid? proof 2

Metin Ersin Arıcan
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    You have one strict increasing sequence $f(x_i)$, but not all strict increasing sequence would go to infinity when $i\to \infty$ – Nick Apr 26 '17 at 16:50
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    That proof assumes $x_i$ converges. – Thomas Andrews Apr 26 '17 at 16:55
  • Why would $f(x_i)$ go to infinity? Even if it doesn't, how do you show that there is a 'maximum point'? – copper.hat Apr 26 '17 at 16:55
  • $f(x_1)=0,f(x_2)=1/2,f(x_3)=2/3,f(x_4)=3/4,\dots$. That sequence increases forever but doesn't go to infinity. – Thomas Andrews Apr 26 '17 at 16:57
  • To prove that a continuous function defined on a compact interval attains its maximum is of fundamental importance to use compactness of the interval. You are not using compactness. – William M. Apr 26 '17 at 17:05
  • Your second proof still assumes the x_i converge which you must prove. THen you state that it is obvious that x_i converge to b which is ... not at all true or obvious. the x_i could converge to halfway between a and b, or any point between. You have a very weak grasp that things increasing don't need to go to infinity or "get pushed to the edge" it's perfectly possible to increase forever and converge to any finite point. – fleablood Apr 26 '17 at 18:13

2 Answers2

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Your proof is not valid.

It is not valid for a few reasons. Firstly, you suggest that if $f(x_1) < f(x_2) < \cdots < f(x_n)$, (and so on for arbitrarily long sequences), then necessarily $f(x_n) \to \infty$. This is not true. In fact, the function $f(x) = x$, where $x_i$ is an increasing sequence converging to $1$, is a counterexample.

There is a deeper reason why your proof is not valid, which is that you presuppose that your sequence of $x_i$ converges to some $x$. This brushes under the rug the most challenging aspects of the classical proofs, which concern the topology of the interval, and in particular concern the compactness of the interval.

davidlowryduda
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  • thanks but i dont understand that "There is a deeper reason why your proof is not valid, which is that you presuppose that your sequence of xi converges to some x". xi is increasing, if xi dont converge to some x, we can find an x > b, isnt it? but it is impossible beacuse f is defined in [a,b]. and in other words it must be converge because for every xi, xi-1 < xi <= b – Metin Ersin Arıcan Apr 26 '17 at 18:28
  • "if xi dont converge to some x, we can find an x > b, isnt it?" Why? We have to prove that. You could have all x_i < k not have them converge to anything. Or can we? Because these are reals and xi <= b there must be a sup {xi}<= b. But does it follow that xi -> sup xi? It does but you must prove that. (Actually, to be honest, I'd give you that one but everyone else who are pointing this out are correct.) – fleablood Apr 26 '17 at 19:21
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The answer by user "mixedmath" as well as various comments amply describe the problems with your proofs. This is a digression which is too long to fit in a comment.

I don't know if you are aware of the book Calculus by M. Spivak. The result you are trying to prove here is one of the Three Hard Theorems. The reason Spivak calls these theorems as hard is because these theorems seem very intuitive yet their proofs are not intuitive at all. In fact any approach based on common sense combined with the high school level knowledge of algebra, calculus is not going to get you a proof.

The proofs of these theorems is based on the completeness property of real numbers and once you know about completeness property these results are trivial applications. But unless you are aware of completeness it is just not possible to prove these. Its good that you tried two proofs of extreme value theorem using your intuition and perhaps realized that one of your proofs was invalid. This kind of exercise will at least help you to understand that these results are not actually that intuitive as they appear to be.

I would suggest you to learn about completeness of real numbers and then you can have a look at the proofs of this theorem in this blog post.

Paramanand Singh
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