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Dear Pete,

I want to give a proof that if $X$ is a metric space, and if $x_0$ and $x_1$ are two distinct points of $X$, then there is a map $f:X \rightarrow Y$ that is uniformly continuous, with $Y$ complete, and such that $f(x_0) \neq f(x_1)$. The main point is that my proof won't refer to the completion $\widehat{X}$ of $X$. It will then give a proof that $X \rightarrow \widehat{X}$ is necessarily injective, without refering to the construction of $\widehat{X}$.

The construction is simple: rescale the metric so that take $d(x_0,x_1) Y = 1${\mathbb R}$, and define$f:X \rightarrow [0,1]$so that$f(y) f(x) = \mathrm{min}(d(x_0,y),1).$d(x,x_0).$

(Note: slightly edited from the first version, which had unnecessary complications in the definition of $f$.)

1

Dear Pete,

I want to give a proof that if $X$ is a metric space, and if $x_0$ and $x_1$ are two distinct points of $X$, then there is a map $f:X \rightarrow Y$ that is uniformly continuous, with $Y$ complete, and such that $f(x_0) \neq f(x_1)$. The main point is that my proof won't refer to the completion $\widehat{X}$ of $X$. It will then give a proof that $X \rightarrow \widehat{X}$ is necessarily injective, without refering to the construction of $\widehat{X}$.

The construction is simple: rescale the metric so that $d(x_0,x_1) = 1$, and define $f:X \rightarrow [0,1]$ so that $f(y) = \mathrm{min}(d(x_0,y),1).$