Let $n$ natural number and $X = [0,1] \times \{1,...,n\}$ with the topology $\tau \times \tau_D$, where $\tau$ is the usual topology of $\mathbb{R}$ induced by $[0,1]$ and $\tau_D$ is the discrete topology in $\{1, . . . , n\}$. We define the equivalence relation of $ (x,n) \sim (x',n')$ iff both points are equal or $x=x'=0.$

My question is, is $X/\sim$ Hausdorff/$T_2$ space? I can't think of how to start, any help is appreciated!

  • 1
    $\begingroup$ Does it help to realize that the quotient space is a hedgehog space with $n$ spikes? Having that visualization may help you prove both that the visualization is correct and that the space is Hausdorff. $\endgroup$ – Brian M. Scott Feb 7 at 22:24

I think you can prove it distinguishing cases:

  • Let $(x,n), (x',n')\in X$ be to points such that $n\neq n'$ and $x,x'\neq0$, then the sets $(0,1]\times \{n\}$ and $(0,1]\times \{n'\}$ are open neighborhoods of each of them respectively (the interval $(0,1]$ is open in $\tau_U|_{[0,1]}$ and the sets $\{n\}$ and $\{n'\}$ are open in $\tau_D$) and their images under the canonic projection of the equivalence relationship are disjoint (again, quite trivially, as the points of the different sets have different second coordinate, and so are not equal, and have a non-$0$ first coordinate, so they can't be in the same equivalence class).

  • Let $(x,n), (x',n)\in X$, with $x\neq x'$. We can assume that $x<x'$ without any loss of generality. Since $\tau_U|_{[0,1]}$ is $T_2$, we can take disjoint neighborhoods of each of them contained in $[0,1]\times \{n\}$ and they will again have images with empty intersection under the projection.

  • The case where $(x,n), (0,n')\in X$, $n\neq n'$ and $x\neq0$ is almost the same as the first one, now taking the neighborhoods $(0,1]\times \{n\}$ and $[0,1]\times \{n'\}$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.