Compactness and continuity.

1. Feb 8, 2008

MathematicalPhysicist

I need to prove that for every continuous function f:X->X of a metric and compact space X, which satisfy for each two different x and y in X p(f(x),f(y))<p(x,y) where p is the metric on X, there's a fixed point, i.e there exist x0 s.t f(x0)=x0.

obviously i thought assuming there isnt such a point i.e that for every x in X f(x)!=x
now because X is compact and it's a metric space it's equivalent to sequence compactness, i.e that for every sequence of X there exist a subsequence of it that converges to x0.

now $$p(f(x_{n_k}),x_{n_k})$$, because they are not equal then there exist e0 such that: $$p(f(x_{n_k}),x_{n_k})>=e0$$
now if x_n_k=f(y_n_k) y_n_k!=x_n_k, we can write it as:
p(x_n_k,x0)+p(x0,y_n_k)>=p(x_n_k,y_n_k)>e0
now if y_n_k were converging to x0, it will be easier, not sure how to procceed...

what do you think?

2. Feb 8, 2008

morphism

This is a special case of the Banach contraction mapping theorem. A proof would go as follows: Let x0 be any point in X, and let xn=f(xn-1) for n>1. Claim: {x_n} converges.

Post back if you need more hints.

3. Feb 9, 2008

MathematicalPhysicist

well cauchy sequnce obviously will do here.
$$p(f(x_n),f(x_{n_k}))<p(x_n,x_n_k)=p(f(x_{n-1}),f(x_{n_k-1}))<p(x_{n-1},x_{n_k-1})<....<p(x_0,x_{n_k-n})$$
now if x_{n_k-n} converges to x_0 (which can be assumed cause it's compact and metric), it will be easy to prove your claim, cause then for every e>0 s.t k is big enough:
p(x_0,x_{n_k-1})<e/2 and also p(x0,x_{n_k-n})<e/2
so p(x_n-1,x0)<=p(f(x_n),f(x_n_k))+p(f(x_n_k),x0)<...<e.

is this wrong? I have a sneaky suspicion that yes.

4. Feb 9, 2008

morphism

If n > n_k, then $x_{n_k - n}$ doesn't make sense. Also I don't see how you can assume that "x_{n_k-n} converges to x_0", because it isn't true.

You started out with the right idea. Let n>m, and consider p(xn, xm). Show that we can make this arbitrarily small. This would imply that {xn} is Cauchy and hence convergent (because X is compact). Then we can use the continuity of f to conclude that f has a fixed point (how?).

5. Feb 11, 2008

MathematicalPhysicist

well, p(xn,xm)<p(xn-1,xm-1)<....<p(xn-m,x0)
now how do i procceed from here?
I mean if I assume n-m is big enough, s.t x_n-m->x0 then that will do, not sure that this is correct...

6. Feb 12, 2008

MathematicalPhysicist

I have another two questions, I need to answer if the next spaces satisfy S2 or Sep, the spaces are with they metrics affiliated with them, in here:
http://www.math.tau.ac.il/~shustin/course/tar5top.xet.pdf
in questions 4,5 (disregard the herbew words near them) there listed the spaces.

well what i think is that because if a space is metric and it satisfies S2 then it also satisifes S2, and always when S2 is satisifes then also Sep is satisifed, then it's easy to check fo Sep, i think it follows that for the first the space follows both of them, while in the second it doesnt satisify either of them.

not sure how argue that?
I mean can I find a countable basis for the C^k[0,1]?
or a countable dense set in it?
what do you think?

7. Feb 16, 2008

morphism

To finish off, you can use the triangle inequality
$$\rho(x_{n-m},x_0) < \rho(x_{n-m},x_{n-m-1}) + \rho(x_{n-m-1}, x_{n-m-2}) + \cdots + \rho(x_1, x_0)$$
coupled with the observation that
$$\rho(x_n, x_{n-1}) < \rho(x_1, x_0)$$.

8. Feb 16, 2008

morphism

As for your other question: I'm guessing S_2 means second countable (has a countable basis) and Sep means separable (has a countable dense subset), right?

And you have the right idea: a metric space is separable iff it's second countable. I would use separability here. For C^k[0,1], try to see if Weierstrauss's theorem is helpful. For l_2, I would think about the subspace consisting of sequences with only finitely many terms. This is certainly dense in l_2, but is it countable? No. So how about we restrict these sequences to those with rational terms?

9. Feb 16, 2008

MathematicalPhysicist

it seems eventually that munkres has a similar questions with hints which were helpful.