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Science Forum Index » Mathematics Forum » metric spaces
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Message |
| Guest |
 Posted: Thu May 01, 2008 12:28 am |
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Let (X, d) be a compact metric space and f : X --> X a self-map such
that
d(f(x), f(y)) = d(x, y) for all x, y in X.
Show that f(X) = X.
(Hint. Suppose there exists x not in f(X), and consider the sequence
{f_n(x)}n>=0, where
f_0(x) = x and fn+1(x) = f(f_n(x)) for all n >= 0. |
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| Someonekicked |
| Posted: Thu May 01, 2008 4:36 am |
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Guest
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On May 1, 1:28 pm, rhvi0...@usyd.edu.au wrote:
Quote: Let (X, d) be a compact metric space and f : X --> X a self-map such
that
d(f(x), f(y)) = d(x, y) for all x, y in X.
Show that f(X) = X.
(Hint. Suppose there exists x not in f(X), and consider the sequence
{f_n(x)}n>=0, where
f_0(x) = x and fn+1(x) = f(f_n(x)) for all n >= 0.
notation will get a complicated, although the idea is "quite" simple.
Now using the hint, you can find a subset {x_n} of f(X) that must
converge lets say to y, then around y we pick elements of {x_n} that
are getting closer to y, and eventually to each others. {x_n} is
special in that the distance between any two elements is equal to the
distance of a third element to x. For every two consecutive elements
that we found around y, there exists a third element getting closer to
x. So gathering all those third elements will give us a sequence that
has x as a limit point, but f(X) is compact, so x must be in f(X),
contradiction. Here are the details:
Clearly f is continuous, so f(X) is compact.
As the hint goes, assume x is in X\f(X), then
Consider the sequence {x_n} for n>= 0, where x_0 = x, x_1 = f(x_0),
x_n = f(x_(n-1))
You can verify that if 1 <= t <= n, then d(f(x_n),x_t) = d(f(x_(n-
t)),x)
Case {x_n} is finite, so for some n>= 0, f(x_n) = x_t, where 0 <= t <=
n.
If t = 0, then f(x_n) = x_0 = x, contradiction.
otherwise, t >= 1, then 0 = d(f(x_n),x_t)= d(f(x_(n - t),x), so f(x_(n
- t) = x, contradiction.
Case {x_n} is infinite,
notice that if i != j, then f(x_i) != f(x_j), or else we will be in
the previous case (*)
Now {x_n} must have at least one limit point in f(X), say y.
For t = 1, there must exist n_1 such that d(x_n_1,y) < 1/t = 1
For t = 2, there must exist n_2, n_2 > n_1 such that d(x_n_2,y) < 1/t
= 1/2
For t = 3, there must exist n_3, n_3 - n_2 > n_2 - n_1 such that
d(x_n_3,y) < 1/t = 1/3
For t>= 3, there must exist n_t, n_t - n_(t-1) > n_(t-1) - n_(t-2)
such that d(x_n_t, y) < 1/t
Now notice that d(x_n_2,x_n_1) < d(x_n_1,y) + d(x_n_2,y) < 1/1 + 1/2
= 3/2
So d(x_n_2, x_n_1) = d(f(x_(n_2 - 1)), x_n_1) = d(f(x_(n_2 - n_1 -
1)), x) < 3/2
In general, d(x_n_(t+1),x_n_t) < d(x_n_(t+1),y) + d(x_n_t,y) < 1/(t+1)
+ 1/t = (2t + 1)/(t^2 + t)
So d(x_n_(t+1),x_n_t) = d(f(x_(n_(t+1)-1)),x_n_t) = d(f(x_(n_(t+1) -
n_t -1)),x) < (2t + 1)/(t^2 + t)
By letting y_1 = f(x_(n_2 - n_1 - 1)), y_t = f(x_(n_(t+1) - n_t -1)),
then we get {y_n} for n in N.
Notice that {y_n} is infinite since we chose n_t's such that
.... > n_(t+1) - n_t > n_t - n_(t-1)> ... > n_2 - n_1 and by (*).
now d(y_t, x) < (2t + 1)/(t^2 + t) -> 0 as t -> +infinity, so {y_t}
converges to x as t -> +infinity
Since f(X) is compact and {y_n} is an infinite subset of f(X), then x
must be in f(X), contradiction! |
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| karl |
| Posted: Thu May 01, 2008 5:29 am |
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Guest
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rhvi0196@usyd.edu.au schrieb:
Quote: Let (X, d) be a compact metric space and f : X --> X a self-map such
that
d(f(x), f(y)) = d(x, y) for all x, y in X.
Show that f(X) = X.
(Hint. Suppose there exists x not in f(X), and consider the sequence
{f_n(x)}n>=0, where
f_0(x) = x and fn+1(x) = f(f_n(x)) for all n >= 0.
Where is your question? |
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