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breaking down the reals

Posted Over 8 Years ago by EN

Here's something I started thinking about while studying modules a few days ago.

Let R be the rest of real numbers, Q be the set of rational numbers, and Z be the set of integers.

Q and R, as abelian groups under addition, can be seen as Z-modules. Q is then a submodule of the Z-module R.

One special type of module is the injective module. There are many equivalent ways to talk about injective modules. See for definitions and examples.

In particular, Q is an injective Z-module. One of the definitions of injective module (the first one on the wiki page) gives us that there exists another submodule K of R such that R is the internal direct sum of Q and K, i.e. Q + K = R and Q intersect K = {0}. Let's talk more about this mystery set K.

Since Q is the set of all rational numbers and the intersection of Q and K is 0, K must contain only irrational numbers (together with 0). Now remember that K is a submodule of R, so it has to be closed under addition. So any of two irrational numbers in K that are not negatives of each other have to add up to give another irrational number in K. This seems pretty strange, since in general the set of irrational numbers is not closed under addition. So K can't be the set of irrational numbers.

Furthermore, since R is uncountable, Q is countable, and Q + K = R, K must be uncountable. Thus K is a proper uncountable subset of the irrational numbers (together with 0) which is closed under addition. Wow.

Since R = Q + K, every real number can be written uniquely in the form q + k, where q is in Q and k is in K. If r is rational, then we have that r = q + 0. If r is irrational, then we have two cases. If r is in K then r = 0 + k. But if r is not in K, then r = q + k for some nonzero q in Q. So some irrational numbers are special in that they have to be written as a nonzero rational plus an irrational.

The set K is unconstructable (go figure) and its existence is based on the Axiom of Choice. We deal with a similar situation when we view R as a vector space over Q. It's been proven (again, using the AoC) that every vector space has a basis. Thus there is a proper subset B of R such that every real number can be written as a unique finite linear combination of rational numbers and numbers from B. This basis must be uncountably infinite for the same reasons as K's, and in fact shares other strange properties with K.

The reals and the Axiom of Choice are bizarre enough on their own. Together they give more interesting things to think about.

There are 4 Replies

That's one of the things that tickled my imagination decades ago when I was enjoying learning about vector spaces. Namely, that even though the real line R looks one-dimensional, it can be viewed as an infinite dimensional space over the field of rationals! (horror)(horror)(horror) In fact it has, as you say, dimension = c, the uncountable cardinal.

So based on that we can construct the complementary submodule K to the rationals inside R. And it works! I don't recall all the details but if I remember correctly we would use the Axiom of Choice in the equivalent form of Zorn's Lemma to get the existence of a maximal independent set which would have to span and be a basis therefore.

So we end up expressing the real line R as an infinite direct sum of copies of the rationals Q

R = ⊕ Q

the infinite direct sum being over the Hamel basis given by Zorn's Lemma. The one thing I don't recall is if we need Cantor's Continuum Hypothesis to do this.

Thanks, great post!

Over 8 Years ago
The Fly

That is quite shocking.

The Axiom of Choice really allows us to do some pretty weird things.

Over 8 Years ago

Could these be used as arguments "against" the Axiom of Choice? {:P}

There are lots of interesting counterintuitive stuff in Math (and today even in Physics). These make the subject quite stimulating and lively! AMAZING! {:D}

Over 8 Years ago
The Fly

[@]EN,Yeano,The Fly:[\@]

One version of the Axiom of Choice is the Russel Multiplicative Principal that:
The Cartesian cross-product of any non empty indexed collection of non empty sets is never empty.
It’s really hard to argue against that.
The Axiom of Choice is not seriously doubted by anyone.

The Continuum Hypothesis is that the first uncountable cardinal number is that of the reals, that is, the cardinal it’s of the powerset of the counting numbers. In other words aleph-1 is 2^ aleph-0.
The CH is much more open to doubt than the AC.

Another version of the AC is that every set can be well-ordered.
This is equivalent to saying that every cardinal number is an aleph. Every cardinal is aleph-x for some ordinal number x.

The Generalized Continuum Hypothesis is that for every cardinal number aleph-x, the next bigger cardinal, namely aleph-(x+1), is the cardinality of the power set of aleph-x, that is, aleph-(x+1) = 2^(aleph-x).
The GCH is even more open to doubt than the CH.

10 Months ago

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