Let $X,Y$ be two compact metric spaces. Suppose there is a sequence of bi-Lipschitz homeomorphisms $f\_n: X\to Y$, and $c\in(0,1]$, satisfying 
$$c\cdot d(x\_1,x\_2)\le d(f(x\_1),f(x\_2))\le \frac{1}{c}\cdot d(x\_1,x\_2),$$
for any $x\_1,x\_2\in X$.


Is is true that one can choose a sub-sequence $f\_{n\_i}$ that uniformly converges to a bi-Lipschitz homeomorphism $f$ (with the same constant $c$)?


Such a statement would imply that in case the Lipschitz distance between two compact metric spaces is bounded, it is actually realised by a bi-Lipschitz homeomorphism (which is better than taking an inf over all bi-Lipschitz homeomorphisms).



Choosing a convergent sub-sequence from a sequence of bi-Lipschitz homeomorphisms
Consider a subsequence $\,g\_n\,$ of $\,f\_n\,$ which is uniformly convergent. Then consider a subsequence $\,h\_n\,$ of $\,g\_n\,$ such that sequence $\,h^{-1}\_n\,$ is uniformly convergent. Then the limit function $\,h\,$ of $\,h\_n\,$ is bilipschitz with the bi-constant equal the same $\,c.$   ***Great***



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> ***REMARK*** This proves what @NikWeaver has already said in his comment under the OP's *Question*. (Yes, I agree with the comment by @NatEldredge).
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