Let $Y\subset\mathbb{R}$ a subspace of the Sorgenfrey line $(\mathbb{R}, \tau_s)$, i.e, $Y$ is a topological space with the same topology $\tau_s$.

Show that $Y$ can't be homeomorph to $\mathbb{R}$ with the usual topology.

I try prove this suposse that yes are homeomorph and find a topological property that satisfied Sorgenfrey line but not $Y$ or viceversa. But i not can't find this property.

  • $\begingroup$ What about $Y=\mathbb{R}$? $\endgroup$ – Vinícius Novelli Nov 5 '17 at 3:29
  • $\begingroup$ @Vinicius Novelli...is for any $Y\subset \mathbb{R}$. $\endgroup$ – Luis Prado Nov 5 '17 at 3:31
  • $\begingroup$ Then it's not true. If $Y=\mathbb{R}$, then it's homeomorphic to Sorgenfrey line. Did you mean a proper subspace? $\endgroup$ – Vinícius Novelli Nov 5 '17 at 3:33
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    $\begingroup$ I feel like I'm missing something. Isn't $\ln:(0,\infty)\rightarrow \mathbb{R}$ a homeomorphism with respect to the Sorgenfrey topology? Because it's monotonic, it maps half closed intervals to half closed intervals..... $\endgroup$ – Jason DeVito Nov 5 '17 at 3:36
  • $\begingroup$ @Vinicius Novelli...yes is a proper subspace...sorry $\endgroup$ – Luis Prado Nov 5 '17 at 3:48

The Sorgenfrey line $S$ is hereditarily disconnected: Any subspace $Y\subset S$ with more than one member is a disconnected space. For if $x,y\in Y$ with $x<y$ then $Y\cap (-\infty,y)$ and $Y\cap [y,\infty)$ are non-empty disjoint OPEN subsets of $Y$ whose union is $Y.$

  • $\begingroup$ The Sorgenfrey line is "strongly zero-dimensional". That is, it has a base (basis) of open-and-closed sets. $\endgroup$ – DanielWainfleet Nov 5 '17 at 20:31

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