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Question about fixed points (metric spaces)

http://www2.imperial.ac.uk/~tsorense/M2PM5/ProblemSheet9.pdf

5(ii). This question looks terribly difficult. Compact implies complete but I can't seem to get anywhere if I try to use a proof similar to that of Banach's fixed point theorem (since we can't construct a Cauchy sequence with that condition). So we need to use a different property of compact spaces. But what? I don't know where to begin.
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nuodai
17
Define a sequence by letting x0∈Xx_0 \in X be any point and xn=f(xn−1)x_n = f(x_{n-1}) for n≥1n \ge 1. As XX is compact, what can you say about this sequence? This will tell you about the existence of a fixed point. For uniqueness, you need to use the fact that d(f(x),f(y))<d(x,y)d(f(x),f(y)) < d(x,y) for x≠yx \ne y.
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gangsta316
OP
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Original post by nuodai
Define a sequence by letting x0∈Xx_0 \in X be any point and xn=f(xn−1)x_n = f(x_{n-1}) for n≥1n \ge 1. As XX is compact, what can you say about this sequence? This will tell you about the existence of a fixed point. For uniqueness, you need to use the fact that d(f(x),f(y))<d(x,y)d(f(x),f(y)) < d(x,y) for x≠yx \ne y.


Convergent subsequence! Ok, but then how do we prove that the limit of this subsequence is a fixed point? Call the limit y. So the subsequence x_n_k -> y, as k-> infinity. We know that since f is continuous (it's uniformly continuous), the sequence f(x_n_k) -> f(y) as k-> infinity. But this doesn't seem to help because if we take f of the subsequence, we may get something that is not in the original subsequence, and then its limit could be anything, not necessarily the same limit y.
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nuodai
17
Original post by gangsta316
Convergent subsequence! Ok, but then how do we prove that the limit of this subsequence is a fixed point? Call the limit y. So the subsequence x_n_k -> y, as k-> infinity. We know that since f is continuous (it's uniformly continuous), the sequence f(x_n_k) -> f(y) as k-> infinity. But this doesn't seem to help because if we take f of the subsequence, we may get something that is not in the original subsequence, and then its limit could be anything, not necessarily the same limit y.


For the sake of simplicity, relabel the xnkx_{n_k} as xkx_k instead (because subscripts of subscripts are ugly). Then by the triangle inequality, d(f(y),y)≤d(f(y),f(xk))+d(xk,y)d(f(y),y) \le d(f(y), f(x_k)) + d(x_k, y), but if k≥Kk \ge K for KK large enough, what is this bounded by? This proves that f(y)=yf(y)=y; next you just need to show that yy is the only point of XX with the property that f(y)=yf(y)=y, which you can deduce using the other inequality.

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