The K-topology on the real line by taking as basis all open intervals $(a,b)$ and $(a,b)\setminus B$ where $K=\{1/n:n=1,2,...\}$ and $B\subset K$. According to this topology,the subset $\mathbb{R}\setminus K$ is open in K-topology but not open in usual topology. I wonder that it can be first countable subspace? thanks in advance


The only point at which the $K$-topology differs locally from the Euclidean topology is $0$: at every $x\in\Bbb R\setminus\{0\}$ the ordinary open intervals around $x$ are a local base at $x$ in both topologies. For $n\in\Bbb Z^+$ let

$$B_n=\left(-\frac1n,\frac1n\right)\setminus K\;;$$

then $\{B_n:n\in\Bbb Z^+\}$ is a countable local base at $0$ in the $K$-topology. Thus, $\Bbb R$ is first countable in the $K$-topology, and so is every subspace.

  • $\begingroup$ This open subspace can be metrizable? $\endgroup$ – ghb Nov 27 '13 at 19:40
  • $\begingroup$ @ghb: The subspace $\Bbb R\setminus K$? Yes: its relative topology is Euclidean. $\endgroup$ – Brian M. Scott Nov 27 '13 at 19:42
  • $\begingroup$ But it is not Euclidean topology, it is finer than Euclidean.I can not see why its relative topology is Euclidean. $\endgroup$ – ghb Nov 27 '13 at 19:45
  • $\begingroup$ @ghb: The $K$-topology is finer than the Euclidean topology; its restriction to $\Bbb R\setminus K$ is the same as the restriction to $\Bbb R\setminus K$ of the Euclidean topology. $\endgroup$ – Brian M. Scott Nov 27 '13 at 19:49
  • $\begingroup$ is it connected? $\endgroup$ – astrobarrel Aug 22 at 17:03

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