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We know that if $T:K\to K$ is a continuous contraction map in a complete metric space $(K,d)$ then it admits an unique fixed point $x$ which is equal to $\lim_{n\to\infty}T^n(z)$ for any $z\in K$.

Does the theorem remain true if we assume that $T$ maps some metric space $(E,d)\supseteq (K,d)$ in $(K,d)$ and we take $z\in E$.

I think that the first assertion (the existence and uniqueness of the fixed point) remains true because it doesn't depend on the point $z$ we take and the second one (the convergence to the fixed point of that limit) remains also true because in the proof of the theorem we only use the fact that some sequence $T^n(y)$ is a Cauchy sequence, but this one lives in a complete metric space.

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Since $T$ maps $E$ into $K$, $Tz\in K$, thus $\{T^nz\}_{n\ge1}\subset K$. Since the sequence is Cauchy and $K$ is complete, the limit is in $K$.

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