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### Equinumerous Sets

```Date: 03/06/2003 at 01:31:27
From: Steve
Subject: Equinumerous Sets

Hi,

I'm trying to show that the set of real numbers and the set of
irrational numbers are the same size.

The main problem is that I don't know how to express elements of
uncountable sets (i.e., I can't enumerate them).  This makes it
difficult to define functions other than those that contain an
explicit formula, such as, e.g., f(x) = x + 1. Also, I can easily
show an injective map from R\Q to R and a surjective map from R to
R\Q, but I get stuck trying to accommodate the remaining elements when
forming a bijection.

If we let R = the reals, Q = the rationals, and R\Q = the irrationals,
I first tried to explicitly set up a bijection between R and R\Q, but
was not successful.  I then split R into the union of disjoint sets
R\Q and Q.  Since Q is denumerable, and since every infinite set
contains a denumerable subset, I've defined a denumerable subset, T,
of R\Q, and clearly there is a bijection from Q to T.  Now, I've also
split up R\Q into the disjoint union of (R\Q)\T and T, so if I can
establish that R\Q and (R\Q)\T are the same size, I'm done.  Was
wondering: (a) if there is an obvious bijection from R to R\Q (or vice
versa), and (b) how I can show that an uncountable set and a proper
(uncountable) subset of that set are the same size (with or without
specifying an explicit bijection). Thanks.  P.S., I'm also trying to
answer this question without making reference to the
Cantor-Schroeder-Bernstein Theorem.
```

```
Date: 03/06/2003 at 03:46:22
From: Doctor Jacques
Subject: Re: Equinumerous Sets

Hi Steve,

First, I would like to point out that your statement (b) is false.

Consider, for example, the sets R and P(R) (the power set of R). These
sets are both uncountable, not equipotent, and the former is a proper
uncountable subset of the latter.

Let us examine the problem at hand.

As Q is countable, we can find an enumeration q_1, q_2, ...

Consider the set T = {eq | q in Q} where e is the well-known constant
2.718... (any irrational would do). T is a subset of the irrationals
(R\Q).

T is also countable; in fact, we have an explicit enumeration of T:

e*q_1, e*q_2, ...

Let S = (R\Q)\T.

We have a partition of R as the disjoint union of Q, T, S, and a
partition of (R\Q) as the disjoint union of T and S.

R     =  Q U T U S
(R\Q) =      T U S

We can first define our bijection f as the identity on S.

It remains to define f as a bijection between (Q union T) and T. As
both sets are countable, and we have an explicit enumeration of them,
this should be easy (think of a bijection between the positive
integers and the positive even integers...)

some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sets

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