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!
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.