Answer to Question #217683 in Differential Geometry | Topology for Prathibha Rose

Question #217683

Prove that seperability is a topological property


1
Expert's answer
2021-07-21T07:26:43-0400

Suppose that XX is a separable space and YY a topological space with f:XYf:X\to Y a homeomorphism between them. Let us prove that YY is separable.

As XX is separable, there exists a countable dense subset DD. We claim that D:=f(D)D':=f(D) is a countable dense subset of YY First of all, as ff is a homeomorphism, it is, in particular, a bijection, so DD' is countable. Secondly, we remark that for a continuous function h:ABh:A\to B we have h1(C)h1(C)\overline{h^{-1}(C)}\subseteq h^{-1}(\overline{C}) for any CBC\subseteq B. Indeed, as hh is continuous, h1(C)h^{-1}(\overline{C}) is a closed set containing h1(C)h^{-1}(C), it contains, in particular, its closure. Finally, we apply this as following : f1(D)f1(D)\overline{f^{-1}(D')} \subseteq f^{-1}(\overline{ D'}), but f1(D)=D=X\overline{f^{-1}(D')}=\overline{D}=X (as ff is bijective, f1(D)=Df^{-1}(D')=D) , so Xf1(D)XX\subseteq f^{-1}(\overline{D'})\subseteq X and therefore f1(D)=Xf^{-1}(\overline{D'})=X. Applying ff to both parts, remembering that it is bijective, we conclude that D=Y\overline{D'}=Y. Therefore, DD' is a countable dense subset of YY, so it is separable.


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