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Ah, he's just getting started on his journey into the set of real numbers! Eventually he will discover, "God made the integers. All else is man made.". In pa
by graycat 13y ago
Ah, he's just getting started on his
journey into the set of real numbers!
Eventually he will discover, "God
made the integers. All else is
man made.".
In particular, man made the real
numbers to be complete which
means that every sequence that
appears to converge, that is,
meets, the Cauchy criterion,
actually does converge. Really
his discoveries are about the
completeness property of the
real numbers. So, in particular,
if we have an infinite series
that meets the Cauchy criterion,
then it converges, in particular,
there is real number for it to
converge to. The same statement
is not true in the rational numbers
or the algebraic numbers!
Calculus: Sure, the elementary properties
of the completeness property of the
real numbers.
How do we know that anything like the
real numbers can exist? Because we
can start with the simplest things,
say, just the empty set, do a
lot of set theory pushing around,
construct something that looks like
the natural numbers -- 1, 2, 3, ....
Then we can use the naturals to
construct (something that looks like)
the integers -- ..., -3, -2, -1, 0,
1, 2, 3, ....
Continuing in this way, we can
construct the rationals and the
reals. For the reals, a popular
approach, nicely intuitive, is
Dedekind cuts.
So, we base it all on just
set theory starting with just
the empty set.
So, the reals exist but only because
man said that they exist!
Too soon he will face a danger,
compactness! That's where
every infinite subset has a limit
point, that is, in the infinite
subset is a sequence that converges
to something. This is true if and
only if every open cover has a
finite subcover. And every
closed and bounded subset of
finite dimensional, real Euclidean space is
compact. A real valued continuous
function with domain a compact
set is uniformly continuous
and bounded and achieves both
its upper and lower bounds.
Seeing these results, the poor
guy might lose it! The usual
way we show that the Riemann
integral of calculus exists is
via uniform continuity.
After he recovers from seeing the
completeness property of the
reals, under no circumstances
let him learn about Hilbert space --
a complete inner product space!
Okay, but are there any examples?
Actually, yes: The set of all
real valued random variables
X so that E[X^2] is finite.
Yup, the set of all of these
is complete and, thus, forms
a Hilbert space. Totally mind
blowing that any such thing
could be true! Here complete
means Cauchy convergent means convergent
where we consider distance
in Hilbert space which is the
metric we get from the
inner product. So, with just
that concept of distance,
a Cauchy convergent sequence of E[X^2] finite
random variables actually
converges, that is, there is
a random variable for the sequence
to converge to.
You'd think that random variables
could wiggle too much, but, no,
they can't! Beyond belief.
If he survives these severe
trials of the mind, keep him
away from a classic text on
point set topology, e.g.,
Kelley. There learn that
can have a set A and a point
x so that point x is right
next to set A but there is
no sequence in set A converging
to point x. That is, sequences
are not enough to characterize
the more general case of convergence.
For this more general case, there
is Moore-Smith convergence,
nets, filters, etc.
Somewhere in there he will discover
the continuum hypothesis, model
theory, etc.
But under no circumstances let him
get near
John C. Oxtoby,
'Measure and Category:
A Survey of the Analogies between
Topological and Measure Spaces',
ISBN 3-540-05349-2,
Springer-Verlag,
Berlin,
1971.
If he sees this book, he may never
recover!
- tome 13y ago> In particular, man made the real numbers to be complete which means that every sequence that appears to converge, that is, meets, the Cauchy criterion, actually does converge. Indeed, but by the same argument as the author's, there are more Cauchy sequences than can be described, so it looks like the real numbers are much bigger than necessary to do mathematics :)
- graycat 13y agoNo, to "do mathematics", e.g., show that the Riemann integral exists, that e and pi exist, etc., we want completeness. Then we are done: The reals are the only complete Archemedean ordered field! So, we have no choice!
- justinpombrio 13y agoOn the contrary, we do have a choice! We could use computable numbers instead of reals. A computable number is any number which is output by some Turing machine, or, equivalently, any number which can be found by some algorithm. e and pi are computable. You are right that Riemann integrals won't exist, but if you modify definitions somewhat, derivatives and integrals can be defined just as easily for computable numbers as for reals and you can do calculus with them (see second link). Unlike reals, computable numbers are countable, and are all describable. (That is, there are aleph-null of them, so there are exactly as many computable numbers as natural numbers, and fewer than real numbers). And while almost all real numbers aren't computable (by the argument in the article), essentially every number you'd ever stumble upon in a math class is. I prefer computable numbers to reals because I have trouble accepting that a thing exists when it by principal cannot be defined. https://en.wikipedia.org/wiki/Computable_number https://en.wikipedia.org/wiki/Computable_number http://www.amazon.com/Computable-Calculus-Oliver-Aberth/dp/0120417529 http://www.amazon.com/Computable-Calculus-Oliver-Aberth/dp/0...
- ColinWright 13y agoNumbers that can be computed by Turing Machines are countable, so list them: c_0, c_1, c_2, etc. Take any interval [a0,b0] (with a0 and b0 computable), and cross out your computable numbers until you find one, say, c_i, in the interval (including an endpoint). Take some strict sub-interval [a1,b1] (again with computable endpoints) within [a0,b0] such that a0<a1, b1<b0, and c_i is not in [a1,b1]. Now continue crossing off computable numbers, each time taking a sub-interval with computable endpoints to exclude any that happen to be in the interval currently under consideration. The left hand end-points, a0, a1, a2, etc, form a strictly increasing sequence of computable points. This sequence is bounded above, and so we would like there to be a least upper bound. And yet it cannot be one of the numbers you first chose, because each has been excluded at some stage. So here we have a strictly increasing, bounded above sequence of computable numbers such that there is no limit in the computable numbers. This proves the computable numbers are not complete, and that makes analysis really messy.