In real analysis, the Bolzano–Weierstrass theorem is a fundamental result about convergence in a finite-dimensional Euclidean space . The theorem states that each bounded sequence in
has a convergent subsequence. An equivalent formulation is that a subset of
is sequentially compact if and only if it is closed and bounded.
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First we prove the theorem in the case of , in which case the ordering on
can be put to good use. Indeed we have the following result.
Lemma: Every sequence in
has a monotone subsequence.
Proof: Let us call a positive integer n a peak of the sequence if implies
, i.e., if
is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks. Then the subsequence of peaks is strictly decreasing, and we are done. So suppose now that there are only finitely many peaks, and let N be the last peak. By definition, this means that there is
with
. Since
,
is not a peak, which, as above, implies the existence of an
with
. Repeating this process leads to an infinite subsequence
Q.E.D
Now suppose we have a bounded sequence in ; by the Lemma there exists a monotone subsequence, necessarily bounded. But it is immediate from the completeness of the real numbers that any bounded nondecreasing (respectively, nonincreasing) sequence converges to its least upper bound (respectively, greatest lower bound).
Finally, the general case can be easily reduced to the case of as follows: given a bounded subsequence in
, 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.
Suppose A is a subset of 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
and a sequence
in A with
for all n, 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
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 is compact if and only if it is closed and bounded. In fact, in a general topology course one learns 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.
The Bolzano–Weierstrass theorem is named after mathematicians Bernard Bolzano and Karl Weierstrass. It was actually first proved by Bolzano, but this proof was lost. It was re-proven by Weierstrass and became an important centerpiece of analysis. Later, it was discovered that Bolzano had in fact proved the theorem long before Weierstrass, hence the current name.
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.