## compactness and continuity.

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?

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 Recognitions: Homework Help Science Advisor 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.
 well cauchy sequnce obviously will do here. $$p(f(x_n),f(x_{n_k}))0 s.t k is big enough: p(x_0,x_{n_k-1}) Recognitions: Homework Help Science Advisor ## compactness and continuity. If n > n_k, then $x_{n_k - n}$ doesn't make sense. Also I don't see how you can assume that "x_{n_k-n} converges to x_0", because it isn't true. You started out with the right idea. Let n>m, and consider p(xn, xm). Show that we can make this arbitrarily small. This would imply that {xn} is Cauchy and hence convergent (because X is compact). Then we can use the continuity of f to conclude that f has a fixed point (how?).  well, p(xn,xm)x0 then that will do, not sure that this is correct...  I have another two questions, I need to answer if the next spaces satisfy S2 or Sep, the spaces are with they metrics affiliated with them, in here: http://www.math.tau.ac.il/~shustin/c...ar5top.xet.pdf in questions 4,5 (disregard the herbew words near them) there listed the spaces. well what i think is that because if a space is metric and it satisfies S2 then it also satisifes S2, and always when S2 is satisifes then also Sep is satisifed, then it's easy to check fo Sep, i think it follows that for the first the space follows both of them, while in the second it doesnt satisify either of them. not sure how argue that? I mean can I find a countable basis for the C^k[0,1]? or a countable dense set in it? what do you think? Recognitions: Homework Help Science Advisor  Quote by loop quantum gravity well, p(xn,xm)x0 then that will do, not sure that this is correct... To finish off, you can use the triangle inequality [tex]\rho(x_{n-m},x_0) < \rho(x_{n-m},x_{n-m-1}) + \rho(x_{n-m-1}, x_{n-m-2}) + \cdots + \rho(x_1, x_0)$$
coupled with the observation that
$$\rho(x_n, x_{n-1}) < \rho(x_1, x_0)$$.

 Recognitions: Homework Help Science Advisor 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?
 it seems eventually that munkres has a similar questions with hints which were helpful.