Bolzano–Weierstrass theorem

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 Rⁿ. The theorem states that each bounded sequence in Rⁿ has a convergent subsequence.^[1] An equivalent formulation is that a subset of Rⁿ is sequentially compact if and only if it is closed and bounded.^[2] The theorem is sometimes called the sequential compactness theorem.^[3]

Proof

First we prove the theorem when $n=1$, in which case the ordering on $\mathbb {R}$ can be put to good use. Indeed, we have the following result.

Lemma: Every sequence $(x_{n})$ in $\mathbb {R}$ has a monotone subsequence.

Proof: Let us call a positive integer $n$ a "peak of the sequence" if $m>n$ implies ${\displaystyle x_{n}>x_{m}}$ i.e., if $x_{n}$ is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks, ${\displaystyle n_{1}<n_{2}<n_{3}<\dots <n_{j}<\dots }$. Then the subsequence ${\displaystyle (x_{n_{j}})}$ corresponding to these peaks is monotonically decreasing. So suppose now that there are only finitely many peaks, let $N$ be the last peak and ${\displaystyle n_{1}=N+1}$. Then $n_{1}$ is not a peak, since ${\displaystyle n_{1}>N}$, which implies the existence of an $n_2>n_1$ with ${\displaystyle x_{n_{2}}\geq x_{n_{1}}}$. Again, ${\displaystyle n_{2}>N}$ is not a peak, hence there is ${\displaystyle n_{3}>n_{2}}$ with ${\displaystyle x_{n_{3}}\geq x_{n_{2}}}$. Repeating this process leads to an infinite non-decreasing subsequence  $x_{n_{1}}\leq x_{n_{2}}\leq x_{n_{3}}\leq \ldots$, as desired.^[4]

Now suppose we have a bounded sequence in $\mathbb {R}$; by the Lemma there exists a monotone subsequence, necessarily bounded. It follows from the monotone convergence theorem that this subsequence must converge.

Finally, the general case can be reduced to the case of $n=1$ as follows: given a bounded sequence in $\mathbb {R} ^{n}$, the sequence of first coordinates is a bounded real sequence, hence has a convergent subsequence. We can then extract a subsubsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence $n$ times — which is still a subsequence of the original sequence — on which each coordinate sequence converges, hence the subsequence itself is convergent.

Alternative proof

There is also an alternative proof of the Bolzano–Weierstrass theorem using nested intervals. We start with a bounded sequence $(x_{n})$:

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  Because ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ is bounded, this sequence has a lower bound $s$ and an upper bound $S$.

• 

  We take ${\displaystyle I_{1}=[s,S]}$ as the first interval for the sequence of nested intervals.

• 

  Then we split $I_{1}$ at the mid into two equally sized subintervals.

• 

  We take this subinterval as the second interval $I_{2}$ of the sequence of nested intervals which contains infinitely many members of ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$. Because each sequence has infinitely many members, there must be at least one subinterval which contains infinitely many members.

• 

  Then we split $I_{2}$ again at the mid into two equally sized subintervals.

• 

  Again we take this subinterval as the third subinterval $I_{3}$ of the sequence of nested intervals, which contains infinitely many members of ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$.

• 

  We continue this process infinitely many times. Thus we get a sequence of nested intervals.

Because we halve the length of an interval at each step the limit of the interval's length is zero. Thus there is a number $x$ which is in each Interval $I_{n}$. Now we show, that $x$ is an accumulation point of $(x_{n})$.

Take a neighbourhood $U$ of $x$. Because the length of the intervals converges to zero, there is an Interval $I_{N}$ which is a subset of $U$. Because $I_{N}$ contains by construction infinitely many members of $(x_{n})$ and ${\displaystyle I_{N}\subseteq U}$, also $U$ contains infinitely many members of $(x_{n})$. This proves, that $x$ is an accumulation point of $(x_{n})$. Thus, there is a subsequence of $(x_{n})$ which converges to $x$.

Sequential compactness in Euclidean spaces

Suppose A is a subset of Rⁿ with the property that every sequence in A has a subsequence converging to an element of A. Then A must be bounded, since otherwise there exists a sequence x_m in A with || x_m || ≥ m for all m, and then every subsequence is unbounded and therefore not convergent. Moreover, A must be closed, since from a noninterior point x in the complement of A one can build an A-valued sequence converging to x. Thus the subsets A of Rⁿ for which every sequence in A has a subsequence converging to an element of A – i.e., the subsets which are sequentially compact in the subspace topology – are precisely the closed and bounded sets.

This form of the theorem makes especially clear the analogy to the Heine–Borel theorem, which asserts that a subset of Rⁿ 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.

History

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.

Application to economics

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 which 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

• Sequentially compact space
• Heine–Borel theorem
• Fundamental axiom of analysis
• Ekeland's variational principle

Notes

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

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

External links

• Hazewinkel, Michiel, ed. (2001), "Bolzano-Weierstrass theorem", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• A proof of the Bolzano–Weierstrass theorem
• PlanetMath: proof of Bolzano–Weierstrass Theorem
• A proof of the Bolzano–Weierstrass theorem as a rap