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Stefan Geschke
Stefan Geschke
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Concerning II and IIa, every subspace of $\mathbb R^n$ that is a complete metric spacecompletely metrizable is in fact a $G_\delta$ set, i.e., a countable intersection of open sets.
If you are not $G_\delta$, you are not homeomorphic to the irrationals.

That completely metrizable subspaces of $\mathbb R^n$ are $G_\delta$ was shown by E. Čech in: On bicompact spaces. Annals of Math. 38 (1937), 823–844.

Concerning II and IIa, every subspace of $\mathbb R^n$ that is a complete metric space is in fact a $G_\delta$ set, i.e., a countable intersection of open sets.
If you are not $G_\delta$, you are not homeomorphic to the irrationals.

That completely metrizable subspaces of $\mathbb R^n$ are $G_\delta$ was shown by E. Čech in: On bicompact spaces. Annals of Math. 38 (1937), 823–844.

Concerning II and IIa, every subspace of $\mathbb R^n$ that is completely metrizable is in fact a $G_\delta$ set, i.e., a countable intersection of open sets.
If you are not $G_\delta$, you are not homeomorphic to the irrationals.

That completely metrizable subspaces of $\mathbb R^n$ are $G_\delta$ was shown by E. Čech in: On bicompact spaces. Annals of Math. 38 (1937), 823–844.

Source Link
Stefan Geschke
Stefan Geschke
  • 16.2k
  • 2
  • 54
  • 82

Concerning II and IIa, every subspace of $\mathbb R^n$ that is a complete metric space is in fact a $G_\delta$ set, i.e., a countable intersection of open sets.
If you are not $G_\delta$, you are not homeomorphic to the irrationals.

That completely metrizable subspaces of $\mathbb R^n$ are $G_\delta$ was shown by E. Čech in: On bicompact spaces. Annals of Math. 38 (1937), 823–844.