§3. Bolzano-Weierstrass and Cauchy's criterion
227
The extension to complex sequences u(n) = v(n) + i.w(n) is immediate:
extract a convergent subsequence v(Pn) from the sequence v(n), then extract
a convergent subsequence from the sequence w(Pn) - not from the complete
sequence wen). After two successive extractions of subsequences the real and
imaginary parts of the new sequence converge, so the sequence does too, qed.
Exercise: prove the BW Theorem directly, using the decimal expansions
of real numbers.
The BW Theorem leads to one of the most fundamental concepts in all
analysis, that of a compact subset of C; this is the term for any set K c C
which is both bounded and closed. The compact sets possess, and are the
only ones to possess, the BW property: every sequence of points of K has a
subsequence converging to a point of K. Proof in two stages:
(i) Every sequence of points of K is, like K, bounded; one can therefore
extract a convergent sequence from it; the limit of the latter belongs to K
just by the definition of closed sets given in nO 1: one cannot escape from K
by passing to a limit.
(ii) Suppose conversely that a set K c C possesses the property in question. First, K is bounded, for if it were not one could, for all n, find an Un E K
such that IUn I > n; it would be impossible to extract a convergent sequence
from the sequence so obtained. On the other hand, K is closed, since if a
sequence u(n) E K converges to a limit u E C, then, by hypothesis, one can
extract a sequence from it which converges to a limit belonging to K; and
this must be u, whence u E K, qed.
The principle of nested intervals extends to every decreasing sequence of
nonempty compact sets Kn C C. We need only choose a u(n) E Kn for all n
and extract a convergent subsequence v(n) = u(Pn) from the sequence u(n).
Since Ki :::> Ki+l :::> ••• and Pi 2: i, we have v(n) E Kp(n) C Kn C Ki for all
n > i, so that the limit of v(n) belongs to each K i , qed.
It is not even really necessary, in the statement above, to assume that
the Kn decrease; it is enough for the partial intersections Kl n K2 n ... n
'Kn = Hn to be nonempty. These decrease and are again compact, since every
intersection, finite or not, of closed sets is again closed, as we saw in nO 1.
Then we apply the preceding result to the Hn.
We shall meet compact sets again later, mainly in the theory of integration. For the moment let us record the following fundamental result:
Theorem 11. Let K be a compact subset of C and let f be a continuous
map of K into C. Then the image f(K) of K under f is compact.
We need only show that f(K) has the BW property. So let yen) be a
sequence of points of f(K); choose x(n) E K such that yen) = f(x(n)); since
K is compact there exists a subsequence x(Pn) which converges to an x E K;
clearly the sequence Y(Pn) then converges to y = f(x) E f(K), qed.
Corollary. Let f be a real function defined and continuous on a compact
subset K of C. Then there are points a, b E K such that
227
The extension to complex sequences u(n) = v(n) + i.w(n) is immediate:
extract a convergent subsequence v(Pn) from the sequence v(n), then extract
a convergent subsequence from the sequence w(Pn) - not from the complete
sequence wen). After two successive extractions of subsequences the real and
imaginary parts of the new sequence converge, so the sequence does too, qed.
Exercise: prove the BW Theorem directly, using the decimal expansions
of real numbers.
The BW Theorem leads to one of the most fundamental concepts in all
analysis, that of a compact subset of C; this is the term for any set K c C
which is both bounded and closed. The compact sets possess, and are the
only ones to possess, the BW property: every sequence of points of K has a
subsequence converging to a point of K. Proof in two stages:
(i) Every sequence of points of K is, like K, bounded; one can therefore
extract a convergent sequence from it; the limit of the latter belongs to K
just by the definition of closed sets given in nO 1: one cannot escape from K
by passing to a limit.
(ii) Suppose conversely that a set K c C possesses the property in question. First, K is bounded, for if it were not one could, for all n, find an Un E K
such that IUn I > n; it would be impossible to extract a convergent sequence
from the sequence so obtained. On the other hand, K is closed, since if a
sequence u(n) E K converges to a limit u E C, then, by hypothesis, one can
extract a sequence from it which converges to a limit belonging to K; and
this must be u, whence u E K, qed.
The principle of nested intervals extends to every decreasing sequence of
nonempty compact sets Kn C C. We need only choose a u(n) E Kn for all n
and extract a convergent subsequence v(n) = u(Pn) from the sequence u(n).
Since Ki :::> Ki+l :::> ••• and Pi 2: i, we have v(n) E Kp(n) C Kn C Ki for all
n > i, so that the limit of v(n) belongs to each K i , qed.
It is not even really necessary, in the statement above, to assume that
the Kn decrease; it is enough for the partial intersections Kl n K2 n ... n
'Kn = Hn to be nonempty. These decrease and are again compact, since every
intersection, finite or not, of closed sets is again closed, as we saw in nO 1.
Then we apply the preceding result to the Hn.
We shall meet compact sets again later, mainly in the theory of integration. For the moment let us record the following fundamental result:
Theorem 11. Let K be a compact subset of C and let f be a continuous
map of K into C. Then the image f(K) of K under f is compact.
We need only show that f(K) has the BW property. So let yen) be a
sequence of points of f(K); choose x(n) E K such that yen) = f(x(n)); since
K is compact there exists a subsequence x(Pn) which converges to an x E K;
clearly the sequence Y(Pn) then converges to y = f(x) E f(K), qed.
Corollary. Let f be a real function defined and continuous on a compact
subset K of C. Then there are points a, b E K such that