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This question is from my topology assignment and I am struck on it.

Give an example of a compact metric space $(X,d)$ , a topological space $(Y,\mathcal{U})$ that is not Hausdorff and a continuous function $f$ that maps $X$ onto $Y$.

Topological space can be taken $Y=\mathbb{N}$ under usual topology. But I am not able to decide $(X,d)$ as compact so that a continuous function exists.

If I take Euclidean metric $d$ then I have to take a closed and bounded subset but I am not able to then find a continuous function then.

Kindly guide!

bof
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  • $\Bbb N$ with the usual topology is Hausdorff. – Brian M. Scott Dec 20 '20 at 19:12
  • Please edit your Question to clean up the wording of the title and notation, spelling, punctuation, etc. in the body. You've asked a couple of hundred Questions here, so it's not unrealistic to expect attention to details. – hardmath Dec 20 '20 at 19:26
  • Why not take any compact metric space $(X,d)$ (with more than one point) and let $Y=X$ with the trivial topology $\mathcal U={\emptyset,Y}$? So $(Y,\mathcal U)$ is a non-Hausdorff topological space, and it shouldn't be hard to find a continuous function that maps $X$ onto $Y$. – bof Dec 21 '20 at 02:41

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HINT: You know that the continuous image of a compact space is compact, so $Y$ must be a non-Hausdorff compact space. An easy way to make sure that $Y$ is compact is to make it finite. We need at least two points in $Y$ if it is to be non-Hausdorff, so let’s let $Y=\{0,1\}$ and $U=\{\varnothing,\{0\},Y\}$. $[0,1]$ is a nice compact metric space; we can try to find a continuous function $f$ mapping $[0,1]$ onto $Y$. What has to happen in order for $f$ to be continuous? We need $f^{-1}[\varnothing]$, $f^{-1}[\{0\}]$, and $f^{-1}[Y]$ to be open in $[0,1]$. Two of those are automatically true; which two? Now find a way to define $f$ so that the third is true as well and $f$ is surjective (onto), and you’ll be done.

Brian M. Scott
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