Recall the contraction mapping theorem:

Let $(X,\rho)$ be a complete metric space, and let $T:X\to X$ and $k\in[0,1)$ be such that$$\rho(T(x),T(y))\leq k\rho(x,y)$$for every $x,y\in X$. Then $T$ has a unique fixed point.

I was taught that if $k=1$, then a fixed point exists, but it may not be unique.

Is this true?

I think it's not since the canonical proof constructs a sequence $\{x_n\}_1^\infty$, where $x_n=T(x_{n-1})$ for $n\geq1$, and where $x_0\in X$ is arbitrary, and then shows that $\{x_n\}$ is Cauchy, which is no longer possible if $k=1$. Do we perhaps require a weaker hypothesis?

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    $\begingroup$ You cannot guarantee a fixed point if $k=1$. Consider $T(x)=x+1$ as a map from $\mathbb R$ to itself. No fixed points but it does preserve distances. $\endgroup$ – User8128 Oct 9 '17 at 18:29
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    $\begingroup$ Even $\rho(T(x),T(y)) < \rho(x,y)$ (for $x \ne y$) is not sufficient for a fixed point: consider $T(x) = x + \text{arccot}(x)$ on $\mathbb R$. $\endgroup$ – Robert Israel Oct 9 '17 at 18:45
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    $\begingroup$ Compactness is not enough either; consider $T(x)=-x$ on $[-2,-1] \cup[1,2]$. Maybe compactness and connectedness is enough? But now you are already getting close to Brouwer which is somehow relatively deep. $\endgroup$ – Ian Oct 9 '17 at 18:53

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