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Roland Bacher
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André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $<x$$< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $>x$$> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).

André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $<x$ in $C$ and where similarly $\beta$ is the infimum of all elements $>x$ in $C$.

André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).

stupid error fixed
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Roland Bacher
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André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{-1/d(x)}$ (extended by$e^{1/(\alpha-x)+1/(x-\beta)}$ if $0$ on$x\not\in C$ where $\alpha$ is the closed set) withsupremum of all elements $d(x)$ denoting$<x$ in $C$ and where similarly $\beta$ is the distanceinfimum of all elements $x$ to the closed set$>x$ in $C$.

André Henriques answer can easily be improved to $C^\infty$ by considering $e^{-1/d(x)}$ (extended by $0$ on the closed set) with $d(x)$ denoting the distance of $x$ to the closed set.

André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $<x$ in $C$ and where similarly $\beta$ is the infimum of all elements $>x$ in $C$.

Source Link
Roland Bacher
  • 17.9k
  • 3
  • 45
  • 113

André Henriques answer can easily be improved to $C^\infty$ by considering $e^{-1/d(x)}$ (extended by $0$ on the closed set) with $d(x)$ denoting the distance of $x$ to the closed set.