**2.1.1. **Show that there is an uncountable number of
Cauchy sequences of rational numbers equivalent to

any given Cauchy sequence of rational numbers.

**Proof.** Let α be any Cauchy sequence of rational numbers, and suppose
that

[ α ] { x l x is a Cauchy sequence and x is equivalent to α }

is countable. We would then have an enumeration of [ α ] , and thus

We define β to be the sequence such that for n ∈ N

Clearly β is equivalent to α, since for each n, β(n) differs from α(n)
by at most . However, we also

have for all n since they differ in the nth
place. Thus β is not in the enumeration of [ α ], and

we have thus obtained a contradiction, so [ α ] is uncountable.

**2.1.2.** Show that every real number can be given by a Cauchy sequence of
rationals , where none

of the rational numbers is an integer.

**Proof.** Let c be an arbitrary real number. If c is an integer, then the
sequence is a

Cauchy sequence of rational numbers that converges to c, none of which are
integers.

Suppose that c is not an integer and let k be an integer such that k - 1 < c <
k. We construct a

sequence by arbitrarily choosing a rational
number from the interval for each
.

Clearly none of the are integers, since c <
< k for all n, and
is equivalent to c, as desired.

**2.1.3**. What kinds of real numbers are representable by Cauchy sequences
of integers.

Only integers are representable as Cauchy sequences of integers. To see this,
let be any

Cauchy sequence of integers. Then there exists
such that j, k ≥ m implies
. If

and were
distinct integers, then would be at least 1,
and thus we must have .

It follows that there exists such that k ≥
m implies , and therefore x is equivalent

to the constant sequence

**2.1.4.** Suppose and
are two sequences of rational numbers.
Define the shuffled sequence

to be Prove that the shuffled sequence is a
Cauchy sequence if and only if

and are equivalent Cauchy sequences.

We will make use of the fact that a sequence is Cauchy if and only if it is
equivalent to some Cauchy

sequence to prove both implications.

**Claim:** If and
are equivalent Cauchy sequences, then the
shuffle sequence

is a Cauchy sequence.

**Proof.** For any ,
there exists such that
implies .

Similarly, since x is equivalent to y, there exists
such that
implies .

Let , and choose k ≥ m. If k is even,
then k = 2k' for some integer

k'. It follows that . It then follows that

by the triangle inequality and since k and k' are greater than both and .

Otherwise, if k is odd, then k = 2k' − 1 for some integer
k'. Similarly, , and thus

since both k and k' are greater than
.

Thus, in either case , so x and z are
equivalent, and z is Cauchy.

**Claim:** If is a Cauchy sequence, then
and
are

equivalent Cauchy sequences.

**Proof.** Since z is Cauchy, for every
there exists m
such that j, k ≥ m implies

. In particular, if we let j = 2k − 1, then
, and we have

for all k ≥ m. Thus, z and x are equivalent, so x is
Cauchy. A similar proof shows that z and

y are equivalent, so y is Cauchy. Finally, since x is equivalent to z and z is
equivalent to y,

x is equivalent to y, by the transitivity of equivalence. Thus if z is Cauchy
then x and y are

both Cauchy and equivalent to each other.

**2.1.8.** Can a Cauchy sequence of positive rational numbers be equivalent
to a Cauchy sequence of negative

rational numbers?

Consider the sequence for
and
for
. Clearly, for any
, we have

, and thus these two sequences are
equivalent.

**2.1.9.** Show that if
is a Cauchy sequence
of rational numbers there exists a positive integer N

such that for all j.

**Proof.** Let x be any Cauchy sequence, and suppose that for every integer N
there exists a j such

that . Since x is Cauchy, there exists m such
that for all k ≥ m,. Let N' be

any integer greater than , and obtain j such
that . If j m, then
,

which is impossible since so
. Thus, j < m. Clearly, there

c an only be finitely many positive j that satisfy this inequality, and thus we
can take the maximum of

. The axiom of Archimedes guarantees that
there is an integer N that is larger than

this maximum, and thus for all j. This is a
contradiction, so there must exist some integer

N such that for all j.