Math for Non-Geeks/The Bolzano-Weierstrass theorem

{{#invoke:Math for Non-Geeks/Seite|oben}}

This article concerns about a theorem that can be useful for many proofs in real analysis (and also in deeper maths topics): the Bolzano-Weierstrass theorem. It is named after Bernard Bolzano and Karl Weierstraß.

This theorem guarantees that there are accumulation points whenever a sequence is bounded. It is crucial for proving existence of limits (or at least accumulation points) for such sequences. One may also carry through these proofs using nested intervals, but it is often shorter to just use the Bolzano-Weierstrass theorem - which already makes use of nested intervals.

Some textbooks use the Bolzano-Weierstrass theorem to prove validity of the monotonicity criterion for sequences and series. One may also go the other way round and prove the Bolzano-Weierstrass theorem if the monotonicity criterion is known to hold. Another implication of the Bolzano-Weierstrass theorem is that continuous functions on a compact interval $[a,b]$ with $a,b\in ℝ$ are bounded and take a minimum and a maximum value.

The Bolzano-Weierstrass theorem

File:Satz von Bolzano-Weierstraß - Quatematik.webm

The Bolzano-Weierstrass theorem reads as follows:

You can intuitively justify this theorem as follows: A sequence is bounded if and only if all of its infinitely many elements fit inside the finite interval $[s,S]$ . Putting infinitely many points in a finite interval will necessarily make it crowded. So there should be some regions with a lot of points, crowding very closely together. The Bolzano-Weierstrass theorem now states that around one point $x$ , there are infinitely many points in each $ϵ$-neighbourhood. No matter how small $ϵ$ is. This $x$ is an accumulation point. Note that $x$ does not need to be part of the sequence and there may be multiple (even overcountably infinitely many) of those $x$.

Why the completeness axiom is necessary

Completeness of the real numbers $ℝ$ means that each Cauchy sequence within $ℝ$ converges to an element in $ℝ$. Roughly speaking, a Cauchy sequence is a sequence which "looks as if it would converge" and it will be precisely defined elsewhere. For now, you need to know that in $ℝ$ and $\mathbb{Q}$, a Cauchy sequence is any sequence with a limit in $ℝ$. Some of these sequences in $\mathbb{Q}$ converges to an element in $ℝ\setminus \mathbb{Q}$, e.g. $(a_n{)}_{n\in \mathbb{N}},a_n={(1+\frac{1}{n})}^n\to e$. The limit $e$ would be the only possible accumulation point of $(a_n{)}_{n\in \mathbb{N}}$. But $e\notin \mathbb{Q}$. So $(a_n{)}_{n\in \mathbb{N}}$ has no limit in $\mathbb{Q}$, which means the Bolzano-Weierstrass theorem cannot hold for $\mathbb{Q}$.

The above example shows that the domain of the sequence elements is crucial for the Bolzano-Weierstrass theorem to hold. The complete $ℝ$ does work, whereas the incomplete $\mathbb{Q}$ does not. Mathematicians often go one step further and give a name to any domain of a sequence, where the Bolzano-Weierstrass theorem works (i.e. Cauchy sequences have a convergent subsequence). Those sets are called compact. For instance, any bounded and closed interval $[s,S]$ is compact. Finite unions of these intervals are compact, too. The real numbers $ℝ$ are not compact, since the Bolzano-Weierstrass theorem only guarantees for an accumulation point if the sequence is bounded within some $[s,S]$. Generally, unbounded sets are not compact.

Proof (alternatively via monotonicity criterion)

A further way to prove the Bolzano-Weierstrass theorem goes by the monotonicity criterion. This criterion says that every monotone and bounded sequence of real numbers converges.

Boundedness is one of the assumptions of the Bolzano Weierstrass. So if we can establish that there is a monotone subsequence, we know that it must converge, which is what we want to show. Can we select such a subsequence? The answer is yes, as the following theorem will prove:

The above Bolzano-Weierstrass selection principle makes it very easy to prove the Bolzano-Weierstrass theorem:

{{#invoke:Math for Non-Geeks/Seite|unten}}