§3. Bolzano-Weierstrass and Cauchy's criterion
229
For all n E N, denote by En the set of up for which p 2: n, i.e. En =
{Un' Un+1' ... }. These En form a decreasing sequence of nonempty subsets
ofR. Condition (1) shows further that the sequence (un), and so the sets En,
are bounded. Let us put
The In are compact intervals.
Since En :J En+1, any number which bounds En above (resp. below) must
also bound En+1 above (resp. below). So
and consequently In :J In+1• The principle of nested intervals then shows the
existence of a U E R belonging to all the In. This is the required limit of the
sequence (un).
For: choose an r > 0 and suppose that (1) is satisfied for p, q > N. By
the definition of EN we then have Ix - yl < r for all x,y E EN, whence
IaN - bNI :::; r since the greatest lower bound and least upper bound of a set
are limits of elements of that set. Since U E IN we thus have Iu - xl :::; r for
all x E IN and in particular for all x E EN. Hence Iu - unl :::; r for n > N,
qed.
Example 1. Consider a closed set F in the plane and, for all x E C, let
d(x, F) = inf d(x, z) = inf Ix - zl
zEF
be the distance from x to F. First, this is a continuous function of x. For all
r > 0 there exists a z E F such that d(x, z) :::; d(x, F) + r, whence, for all
11 E C,
d(y, z) :::; d(y, x) + d(x, F) + r
and so d(y, F) :::; d(y, x)+d(x, F)+rj since r is arbitrary one may deduce that
d(y, F) :::; d(x, F) + d(y, x) and also the inequality obtained by interchanging
x and Yj whence finally the relation
Id(x,F) - d(y,F)1 :::; d(x,y)
which proves continuity, whether F is closed or not.
When F is closed then for any x there exists at least one a E F for which
d(x, a) = d(x, F), i.e. a point at the minimum distance to F. This is obvious
if d(x, F) = 0: there are then an E F such that d(an , x) tends to 0, so that
x is adherent to F and so belongs to F. If d(x, F) = d > 0, consider the
intersection K of F and the closed ball B(x,2d)j this is a closed bounded
set, so compact. By the definition of a greatest lower bound, for any n there
229
For all n E N, denote by En the set of up for which p 2: n, i.e. En =
{Un' Un+1' ... }. These En form a decreasing sequence of nonempty subsets
ofR. Condition (1) shows further that the sequence (un), and so the sets En,
are bounded. Let us put
The In are compact intervals.
Since En :J En+1, any number which bounds En above (resp. below) must
also bound En+1 above (resp. below). So
and consequently In :J In+1• The principle of nested intervals then shows the
existence of a U E R belonging to all the In. This is the required limit of the
sequence (un).
For: choose an r > 0 and suppose that (1) is satisfied for p, q > N. By
the definition of EN we then have Ix - yl < r for all x,y E EN, whence
IaN - bNI :::; r since the greatest lower bound and least upper bound of a set
are limits of elements of that set. Since U E IN we thus have Iu - xl :::; r for
all x E IN and in particular for all x E EN. Hence Iu - unl :::; r for n > N,
qed.
Example 1. Consider a closed set F in the plane and, for all x E C, let
d(x, F) = inf d(x, z) = inf Ix - zl
zEF
be the distance from x to F. First, this is a continuous function of x. For all
r > 0 there exists a z E F such that d(x, z) :::; d(x, F) + r, whence, for all
11 E C,
d(y, z) :::; d(y, x) + d(x, F) + r
and so d(y, F) :::; d(y, x)+d(x, F)+rj since r is arbitrary one may deduce that
d(y, F) :::; d(x, F) + d(y, x) and also the inequality obtained by interchanging
x and Yj whence finally the relation
Id(x,F) - d(y,F)1 :::; d(x,y)
which proves continuity, whether F is closed or not.
When F is closed then for any x there exists at least one a E F for which
d(x, a) = d(x, F), i.e. a point at the minimum distance to F. This is obvious
if d(x, F) = 0: there are then an E F such that d(an , x) tends to 0, so that
x is adherent to F and so belongs to F. If d(x, F) = d > 0, consider the
intersection K of F and the closed ball B(x,2d)j this is a closed bounded
set, so compact. By the definition of a greatest lower bound, for any n there