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Joseph Van Name
Joseph Van Name
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Such a map does not exist since a continuous map on $X$ must be constant almost everywhere. In other words, if $f:X\rightarrow X$ is a map such that $f^{-1}[A]\in\mathcal{U}$$f^{-1}[A]\in u$ for each $A\in\mathcal{U}$$A\in u$, then $\{x\in X|f(x)=x\}\in\mathcal{U}$$\{x\in X|f(x)=x\}\in u$. For a proof, see the book The Theory of Ultrafilters by Comfort and Negrepontis Thm 9.2 or Andreas Blass's dissertation p. 12.

Such a map does not exist since a continuous map on $X$ must be constant almost everywhere. In other words, if $f:X\rightarrow X$ is a map such that $f^{-1}[A]\in\mathcal{U}$ for each $A\in\mathcal{U}$, then $\{x\in X|f(x)=x\}\in\mathcal{U}$. For a proof, see the book The Theory of Ultrafilters by Comfort and Negrepontis Thm 9.2 or Andreas Blass's dissertation p. 12.

Such a map does not exist since a continuous map on $X$ must be constant almost everywhere. In other words, if $f:X\rightarrow X$ is a map such that $f^{-1}[A]\in u$ for each $A\in u$, then $\{x\in X|f(x)=x\}\in u$. For a proof, see the book The Theory of Ultrafilters by Comfort and Negrepontis Thm 9.2 or Andreas Blass's dissertation p. 12.

Source Link
Joseph Van Name
Joseph Van Name
  • 1
  • 4
  • 75
  • 118

Such a map does not exist since a continuous map on $X$ must be constant almost everywhere. In other words, if $f:X\rightarrow X$ is a map such that $f^{-1}[A]\in\mathcal{U}$ for each $A\in\mathcal{U}$, then $\{x\in X|f(x)=x\}\in\mathcal{U}$. For a proof, see the book The Theory of Ultrafilters by Comfort and Negrepontis Thm 9.2 or Andreas Blass's dissertation p. 12.