= Strictly distance-decreasing map
{title2=$d(Tx,Ty)<d(x,y)\quad(x\ne y)$}
= Strictly distance-decreasing
{synonym}
A <strictly distance-decreasing map> decreases the distance between every pair of distinct points. Unlike a <contraction mapping>, it need not have a single <Lipschitz constant> below one. Two distinct <fixed points> would contradict strict decrease, so there is at most one. Existence can fail even on a <complete metric space>: on $[0,\infty)$, $T(x)=x+1/(1+x)$ has no <fixed point>, while the <mean value theorem> and $0\le T'(x)<1$ show strict distance decrease.
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