Bolzano-Weierstrass Theorem
Appendices, Volume 3, Why Slopes and More Math.
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, \ldots) with the property that $j\ne
k$ implies $q_j \ne q_k$. 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 $s\ne A$ of the set $S$.
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[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 $[a,b]$.
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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 $\frac12 10^{-(k+1)}$ 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 $\frac1210^{-m}$
about $L$, there are infinitely many points of the infinite set $S$.
Details follow.
\noindent \proofsize 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 $0.c_1c_2\ldots c_k$
which coincides with the first $k$ places of the decimal expansion
$0.c_1c_2\ldots c_kc_{k+1}$ of the left-end of the next subinterval.
This process yields an infinite decimal expansion $0.c_1c_2c_3\ldots$.
This defines a real number $L=0.c_1c_2c_3\ldots$ 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 $B\le A$ is another limit
point of the set $S$. Then $A-B>10\cdot 10^{-k}$ for some whole
number $k$. Further for such a whole number $k$, all intervals of
length $2\cdot10^{-m}<10^{-k}$ centered at $B$ would also contain a
point of $S$. And the latter would be imply there was an interval $J_k$
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 $A_k=c_1c_2\ldots c_p.a_1a_2\ldots a_k$, has the
property that the interval $I_k=[A_k,A_k+10^{-k}]$ is the leftmost
interval with infinitely many points of $S$. Yet $J_k$ is to the left
of $I_k$. This is a contradiction. Thus the supposition that there
exists a limit point $B$ of $S$ with $B
Note the above arguments or reasoning depends
on the assumption that an infinite decimal expansions yields a real
number. Base two or any other base $m\ge2$ could have been used instead.
The selection of base ten in the above argument is a historical and
cultural preference.
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 $k\ge 1$. 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.
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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