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.(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]p(f(x_{n_k}),x_{n_k})[/tex], because they are not equal then there exist e0 such that: [tex]p(f(x_{n_k}),x_{n_k})>=e0[/tex]

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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# Compactness and continuity.

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