We have to decide if there is a local homeomorphism between $\mathbb{R}^{4}$ and (X$\times$Z), where X=$\mathbb{C}^{2}$ with metric topology and Z=$\mathbb{C}^{2}$ with cofinite topology. Here is my attempt at it:

If there was a local homeomorphism between g: $\mathbb{R}^{4}$$\rightarrow$(X$\times$Z) then for x$_{0}$ in $\mathbb{R}^{4}$, there will be an open set U $\subseteq$$\mathbb{R}^{4}$ such that g(U) is homeomorphic to U. However, U in metric topology is Hausdorff whereas g(U)$\subseteq$(X$\times$Z) can not be Hausdorff because (X$\times$Z) is not Hausdorff.

I also have a separate proof that since Z is not Hausdorff in cofinite topology, (X$\times$Z) is not Hausdorff. I would really appreciate any comments on this.

  • 3
    $\begingroup$ correct correct correct $\endgroup$ – Forever Mozart Apr 28 '16 at 4:55
  • $\begingroup$ thank you very much @foreverMozart $\endgroup$ – HumbleStudent Apr 28 '16 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.