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Bolzano-Weierstrass Theorem

A set S is said to be infinite if and only if it contains an infinite sequence of points q_j (j=1,2,3, ...) with the property that j ¹ k implies q_j ¹ q_k. A real number A is said to be a limit point of a infinite set of numbers S if and only if every interval I centered at A, no matter how small, must contain infinitely many elements of the set S. Equivalently, a real number A is said to be a limit point of a set S if and only if every interval I centered at A contains at least one element s ¹ A of the set S.

Theorem B.1 [Bolzano-Weierstrass] If an infinite set S is contained in a finite interval [a,b] then the set S has at least one limit point A in the interval [a,b].

 The main idea in the proof given next is as follows. For each whole number k, in the finite interval [a,b] there must be a leftmost subinterval I_k of length 10^-k which contains infinitely many points, which in turn must contain an leftmost interval I_k+1 of length [1/2] 10^-(k+1) which contains infinitely many points, and so on. It follows that the left end points of these nested (each inside its predecessor) intervals form a Cauchy sequence with a limit L. This limit has the property that in every interval of length [1/2]10^-m about L, there are infinitely many points of the infinite set S. Details follow.

Proof of the Bolzano-Weierstrass Theorem.

For each whole number k > 0, the interval [a,b] is covered by subintervals, each of which has length 10^-k. We can choose these subintervals so that their end points have a decimal representation that ends k places after the decimal point. If all the subintervals contain only finitely many set members, the set S would be finite. So at least one of the subintervals must have infinitely many elements of S. While we don't have enough information to say which of the finitely many subintervals must has them, there must be at least one subinterval. Choose one, say the leftmost one, and call it I_k.

Now we have a sequence of intervals I_k. The above choice implies that I_k+1 is contained within I_k. The reason for this follows - a proof within a proof: The interval I_k+1 must be to the left, to the right, or inside of the interval I_k. In the first case, I_k+1 would lie in an interval of length 10^-k which is to left of I_k and has infinitely points, in particular, those in I_k+1. But the latter is impossible because of the leftmost selection of I_k. In the second case, the interval I_k contains 10 subintervals of length 10^-(k+1). But if I_k+1 is to the right of the interval I_k then each of these ten subintervals contains only finitely many points of the set. This contradicts the choice of I_k. The only possibility that remains is that I_k+1 is contained within I_k. That completes this proof within a proof.

The foregoing selection of intervals I_k yields, a nested collection of subintervals, each of length 10^-k and each of which contains infinitely many elements of the original set.

End of proof. The above selection process implies that the left end-point of the k-th subinterval has a finite decimal expansion 0.c₁c₂¼c_k which coincides with the first k places of the decimal expansion 0.c₁c₂¼c_kc_k+1 of the left-end of the next subinterval. This process yields an infinite decimal expansion 0.c₁c₂c₃¼. This defines a real number L = 0.c₁c₂c₃¼ with the property that each neighborhood interval [L-10^-k,L+10^k] contains infinitely many elements of the original set.

The conclusion (or assumption) that a bounded infinite set of real numbers has a limit point has many consequences in arithmetic-based mathematics. The above proof identifies the leftmost limit point of the set S.

Here for the sake of contradiction, suppose B < A is another limit point of the set S. Then A-B > 10·10-k for some whole number k. Further for such a whole number k, all intervals of length 2·10-m < 10-k centered at B would also contain a point of S. And the latter would be imply there was an interval Jk of length 10-k containing B and to the left of A with infinitely many points of S. But the decimal expansion of A to k-decimal places, say Ak = c1c2¼cp.a1a2¼ak, has the property that the interval Ik = [Ak,Ak+10-k] is the leftmost interval with infinitely many points of S. Yet Jk is to the left of Ik. This is a contradiction. Thus the supposition that there exists a limit point B of S with B < A must be false.

FOOTNOTE: The set-theoretic formulation of modern math moves away from this preference.

The above argument, that is proof, relies on the in principle ability to choose subintervals, one inside another, repeatedly, one for each integer k ³ 1. The above demonstration (proof) is appealing, but some could object to the use in-principle part of this argument. There is no practical or constructive way to make the choice. That is, the choice is possible in principle, but not in practice. For each whole number k, the information that the set S is infinite is insufficient by itself to identify in practice the leftmost interval of length 10^-k with an infinite number of points in S.

The above construction of a nested sequence of intervals is a plausible argument for some, but only a figment of the imagination for others. There is a division among mathematicians on whether or not thought-based but impractical choice-based existence arguments (or constructions) are acceptable. The most rigorous and also the most limited perspective is that the above kind of argument is heuristic and somewhat plausible, but not reliable. Another perspective is that the above argument is acceptable and reliable. Suffice it to say that direct, not just in principle, existence proofs are more welcome and more certain in the mathematical reasoning process than other kinds of proof. However, some of the following theorems depend on the above theorem and hence the above in-principle construction.

FOOTNOTE: More Food for thought: Compare and contrast the role of choice in the above argument with the role of Maxwell's Demon in improbable gas dynamics. There may be a limit to what is acceptable in principle as a conclusion-reaching method.

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