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I came across the following version of Bolzano-Weierstrass theorem:

Let a function $f:[a,b]\rightarrow \Bbb R$ be continuous on the segment $[a,b]\subseteq \Bbb R$. Then $f([a,b])=[m,M]$ is also a segment.

Can someone explain what this means(preferably with an example)? Why is this theorem important?

lmc
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    It simply means that $f$ preserves connectedness. In other words, first observe that $[a,b]$ is compact so $f$ achieves a maximum $M$ and a minimum $m$. Now try proving that $\forall m \leq c \leq M, \exists x$ such that $f(x) =c$ – kishlaya Feb 12 '17 at 08:58
  • This means that the function takes its maximum en minimum somewhere. By the continuity and the intermediate value theorem it also takes every value in between. – Shashi Feb 12 '17 at 08:59
  • And this doesn't look like BW theorem, but rather like a simple use of the Intermediate Value Theorem. – DonAntonio Feb 12 '17 at 09:14
  • This is also known as https://en.wikipedia.org/wiki/Extreme_value_theorem – rtybase Feb 12 '17 at 09:43
  • @DonAntonio ... Intermediate value theorem tells you the range is some interval. But it does not tell you that the range is bounded, or (even if it is) that the range includes the endpoints $m, M$. – GEdgar Feb 12 '17 at 13:32
  • @GEdgar Fair enough, then IVT + Weierstrass Theorems I-II , but not BW Theorem which tells you that a bounded infinite sequence has a convergent subsequence. – DonAntonio Feb 12 '17 at 14:24

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There are three crucial steps:

1) Bolzano-Weierstrass Theorem stating that

Every bounded sequence of real numbers has a convergent subsequence.

The method of the proof sometimes is jokingly referred to as "how to catch the lion in the Sahara desert".

2) Next, is the First Weierstrass theorem stating that

Every function $f$ which is continuous on a (finite) closed interval $[a,b]$ of $\mathbf R$ is bounded.

The standard proof is towards a contradiction: suppose that there is a sequence $(x_n)$ of points of $[a,b]$ such that $$ |f(x_n)| \to +\infty. $$ Then we shall quickly obtain a contradiction if we apply the Bolzano--Weierstrass theorem (think how it may be done).

3) Finally, the Second Weierstrass theorem a.k.a. the Extreme Value Theorem stating that

Every function $f$ which is continuous on a (finite) closed interval $[a,b]$ of $\mathbf R$ attains in this interval its minimum and its maximum value.

To decipher, there are $x_0,x_1 \in [a,b]$ such that $$ m=f(x_0) \le f(x) \le f(x_1)=M $$ for all $x \in [a,b]$ (in particular, the image of $[a,b]$ under $f$ is $[m,M]$). Naturally, we apply the First Weierstrass theorem.

Studying examples to understand the Second Weierstrass Theorem would be of a little help; concepts and proofs matter more.

Olod
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  • Great answer! (+1) –  Feb 12 '17 at 10:33
  • Don't you also need to use the Intermediate Value Theorem to know that every element of $[m,M]$ is actually attained as a value of $f$? – tracing Feb 12 '17 at 12:24
  • Certainly (among many other things, like completeness). We were asked to provide some background to understand the statement, you can't mention everything in a short answer. – Olod Feb 12 '17 at 12:39
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This is not exactly Bolzano Weierstrass theorem. Bolzano Weierstrass theorem has two forms:

  • Any infinite bounded subset of real numbers has an accumulation point.
  • Any bounded sequence has a convergent subsequence.

You can see that both these forms of Bolzano Weierstrass are equivalent. The theorem mentioned in question combines three famous properties of continuous functions on a closed interval. These are what Michael Spivak calls Three Hard Theorems in his book Calculus. Needless to say these theorems are of theoretical nature and are easy consequences of the completeness property of real numbers. Spivak calls them hard only because these are not proved in most calculus textbooks. These I state below:

Combining these three properties we can easily see that the range of a function $f$ continuous on $[a, b] $ is also a closed interval (first two properties ensure that there is a minimum and a maximum value of $f$ and the third one ensures that $f$ takes all the values between these two extreme values).

  • I don't know man, my textbook says that's Bolzano-Weierstrass theorem, and the one you mentioned that any bounded sequence has a convergent sequence is called Weirestrass theorem. – lmc Feb 12 '17 at 13:11
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    @Lewis: maybe your textbook is different from majority of textbooks out there. And it does not matter. You need to stick to the book prescribed by your teacher/professor because you will be graded on the basis of that in exam. – Paramanand Singh Feb 12 '17 at 13:21