Bolzano–Weierstrass theorem

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In mathematics, specifically in real analysis, the Bolzano–Weierstrass theorem, named after Bernard Bolzano and Karl Weierstrass, is a fundamental result about convergence in a finite-dimensional Euclidean space . The theorem states that each infinite bounded sequence in has a convergent subsequence.[1] An equivalent formulation is that a subset of is sequentially compact if and only if it is closed and bounded.[2] The theorem is sometimes called the sequential compactness theorem.[3]

History and significance

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The Bolzano–Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. It was actually first proved by Bolzano in 1817 as a lemma in the proof of the intermediate value theorem. Some fifty years later the result was identified as significant in its own right, and proved again by Weierstrass. It has since become an essential theorem of analysis.

Proof

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First we prove the theorem for   (set of all real numbers), in which case the ordering on   can be put to good use. Indeed, we have the following result:

Lemma: Every infinite sequence   in   has an infinite monotone subsequence (a subsequence that is either non-decreasing or non-increasing).

Proof[4]: Let us call a positive integer-valued index   of a sequence a "peak" of the sequence when   for every  . Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices   and the following terms  . So, the infinite sequence   in   has a monotone (non-increasing) subsequence, which is  . But suppose now that there are only finitely many peaks, let   be the final peak if one exists (let   otherwise) and let the first index of a new subsequence   be set to  . Then   is not a peak, since   comes after the final peak, which implies the existence of   with   and  . Again,   comes after the final peak, hence there is an   where   with  . Repeating this process leads to an infinite non-decreasing subsequence   , thereby proving that every infinite sequence   in   has a monotone subsequence.

Now suppose one has a bounded sequence in  ; by the lemma proven above there exists a monotone subsequence, likewise also bounded. It follows from the monotone convergence theorem that this subsequence converges.

The general case ( ) can be reduced to the case of  . Firstly, we will acknowledge that a sequence   (in   or  ) has a convergent subsequence if and only if there exists a countable set   where   is the index set of the sequence such that   converges. Let   be any bounded sequence in   and denote its index set by  . The sequence   may be expressed as an n-tuple of sequences in   such that   where   is a sequence for  . Since   is bounded,   is also bounded for  . It follows then by the lemma that   has a convergent subsequence and hence there exists a countable set   such that   converges. For the sequence  , by applying the lemma once again there exists a countable set   such that   converges and hence   has a convergent subsequence. This reasoning may be applied until we obtain a countable set   for which   converges for  . Hence,   converges and therefore since   was arbitrary, any bounded sequence in   has a convergent subsequence.

Alternative proof

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There is also an alternative proof of the Bolzano–Weierstrass theorem using nested intervals. We start with a bounded sequence  :

Because we halve the length of an interval at each step, the limit of the interval's length is zero. Also, by the nested intervals theorem, which states that if each   is a closed and bounded interval, say

 

with

 

then under the assumption of nesting, the intersection of the   is not empty. Thus there is a number   that is in each interval  . Now we show, that   is an accumulation point of  .

Take a neighbourhood   of  . Because the length of the intervals converges to zero, there is an interval   that is a subset of  . Because   contains by construction infinitely many members of   and  , also   contains infinitely many members of  . This proves that   is an accumulation point of  . Thus, there is a subsequence of   that converges to  .

Sequential compactness in Euclidean spaces

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Definition: A set   is sequentially compact if every sequence   in   has a convergent subsequence converging to an element of  .

Theorem:   is sequentially compact if and only if   is closed and bounded.

Proof: (sequential compactness implies closed and bounded)

Suppose   is a subset of   with the property that every sequence in   has a subsequence converging to an element of  . Then   must be bounded, since otherwise the following unbounded sequence   can be constructed. For every  , define   to be any arbitrary point such that  . Then, every subsequence of   is unbounded and therefore not convergent. Moreover,   must be closed, since any limit point of  , which has a sequence of points in   converging to itself, must also lie in  .

Proof: (closed and bounded implies sequential compactness)

Since   is bounded, any sequence   is also bounded. From the Bolzano-Weierstrass theorem,   contains a subsequence converging to some point  . Since   is a limit point of   and   is a closed set,   must be an element of  .

Thus the subsets   of   for which every sequence in A has a subsequence converging to an element of   – i.e., the subsets that are sequentially compact in the subspace topology – are precisely the closed and bounded subsets.

This form of the theorem makes especially clear the analogy to the Heine–Borel theorem, which asserts that a subset of   is compact if and only if it is closed and bounded. In fact, general topology tells us that a metrizable space is compact if and only if it is sequentially compact, so that the Bolzano–Weierstrass and Heine–Borel theorems are essentially the same.

Application to economics

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There are different important equilibrium concepts in economics, the proofs of the existence of which often require variations of the Bolzano–Weierstrass theorem. One example is the existence of a Pareto efficient allocation. An allocation is a matrix of consumption bundles for agents in an economy, and an allocation is Pareto efficient if no change can be made to it that makes no agent worse off and at least one agent better off (here rows of the allocation matrix must be rankable by a preference relation). The Bolzano–Weierstrass theorem allows one to prove that if the set of allocations is compact and non-empty, then the system has a Pareto-efficient allocation.

See also

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Notes

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  1. ^ Bartle and Sherbert 2000, p. 78 (for R).
  2. ^ Fitzpatrick 2006, p. 52 (for R), p. 300 (for Rn).
  3. ^ Fitzpatrick 2006, p. xiv.
  4. ^ Bartle and Sherbert 2000, pp. 78-79.

References

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  • Bartle, Robert G.; Sherbert, Donald R. (2000). Introduction to Real Analysis (3rd ed.). New York: J. Wiley. ISBN 9780471321484.
  • Fitzpatrick, Patrick M. (2006). Advanced Calculus (2nd ed.). Belmont, CA: Thomson Brooks/Cole. ISBN 0-534-37603-7.
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