Exam (elaborations)
METRIC SPACES D – COMPLETENESS AND CONNECTEDNESS
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METRIC SPACES D – COMPLETENESS AND CONNECTEDNESS...
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METRIC SPACES D – COMPLETENESSAND CONNECTEDNESS
Cauchy sequence, definition - ANSWER (x_n) a sequence in a metric space X
is Cauchy if for all e>0 there exists N st. n, m >= N implies d(x_n, x_m) < e.
Convergent sequence, definition - ANSWER (x_n) a sequence in a metric
space X is Cauchy if there exists a in X such that for all e>0 there exists N st. n
>= N implies d(x_n, a) < e.
Convergent sequences, Cauchy sequences and bounded sequences - ANSWER
Convergent sequences are Cauchy, Cauchy sequences are bounded sequences,
and the converses are in general false
Completeness, definition and examples - ANSWER X a metric space is
complete if every Cauchy sequence converges.
R is complete; (-1, 1) is not complete, but tan(pix) is a homeomorphism from
(0, 1) to R, so completeness is not preserved by homeomorphisms.
Any set with a discrete metric is complete; every set is a subspace of a complete
space.
(B(X), ||.||_inf) is complete always, B(X) the set of bounded functions on X.
Completeness of subspaces - ANSWER Y in complete X is complete iff Y is
closed in X, as:
If Y is not closed and X is complete, there exists (y_n) in Y st. y_n -> x in X\Y
(as X is complete). Then (y_n) is Cauchy in X, so (y_n) is Cauchy in Y despite
not converging in Y, so Y is not complete.
On the other hand, if Y is closed, suppose y_n is Cauchy in Y; then it is Cauchy
in X, so it converges to some x in X. Then as Y is closed, x is in Y always. So Y
is complete.
, Completeness of l^p spaces, statement and proof - ANSWER l^1, l^2, ... l^inf
are complete spaces, as:
jesus, **** this proof
Completeness of L^p spaces, statement - ANSWER they are
Lipschitz function, definition - ANSWER f: (X, d_X) -> (Y, d_Y) is Lipschitz
if there exists a K st. for all x1, x2 in X:
d_Y(fx1, fx2) =< K d_X(x1, x2).
Contraction definition - ANSWER f: (X, d_X) -> (Y, d_Y) is a contraction if
there exists a K < 1 st. for all x1, x2 in X:
d_Y(fx1, fx2) =< K d_X(x1, x2).
Contraction mapping theorem, statement - ANSWER X a nonempty complete
metric space and f: X -> X a contraction; then there exists a unique x in X st.
f(x) = x.
Contraction mapping theorem, proof - ANSWER if there were two points x1
and x2 such that f(x1) = x1 and f(x2) = x2, then d(x1, x2) = d(f(x1), f(x2)) =<
Kd(x1, x2) so 1 =< Kd(x1, x2), contradiction, unless d(x1, x2) = 0, ie. x1 = x2.
on the other hand, x st. f(x) = x exists, as take any x_0 and let x_n+1 = f(x_n);
then d(x_(n-1), x_n) =< K^(n-1)d(x_0, x_1) clearly. Let d(x_0, x_1) = D.
So with m > n:
d(x_n, x_m) =< d(x_n, x_(n+1)) + ... + d(x_(m-1), x_m)
=< (K^n)D + ... + K^(m-1)D
=< DK^n (1 + K + ...) = DK^n/(1-K)
so taking n to infinity, d(x_n, x_m) goes to 0, so x_n is Cauchy, so x_n
converges to some x as X is complete.
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