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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Let $overlineX rightarrow X$ be a covering space, where $overlineX$ is a Hausdorff space. Does $X$ have to be Hausdorff? I assume there is a counterexample, but I couldn't find one yet. general-topology covering-spaces separation-axioms share | cite | improve this question edited Jul 14 at 22:00 Eric Wofsey 163k 12 189 301 asked Jul 14 at 20:34 L. Gitin 58 4 Try $X$ nonHausdorff and let the cover be a product with a discrete space. – Randall Jul 14 at 21:00 1 Why is $bar X$ Hausdorff then? – Fan Zheng Jul 14 at 21:02 You’re right that doesn’t work. Never mind. – Randall Jul 14 at 21:27 1 Example here. –  user574380 Jul 14 at 21:33 1 It is true if the fibres $p^-1(x), x in X$ are finite. See math.stackexchange.com/q/396989. – Paul Frost Jul 14 at 21:39 add a comment  | 

Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite Basically I'm trying to prove a limit with the delta epsilon definition and I'm reading about how to do it. In one example, the book does the following: $lim_(x,y)to (1,-1) fracy+1(3(x-1)^2+2(y+1)^2)^1/3 leq lim_(x,y)to (1,-1) fraclVert[x-1,y+1]rVert(2lVert[x-1,y+1]rVert ^2+2lVert[x-1,y+1]rVert^2)^1/3$ I don't understand why this is true. I know for a fact that $(y+1) leq ||[x-1,y+1]||$ but I don't understand what's going on in the denominator. As far as I know the denominator of the first function should be greater than or equal to the one on the second function for this inequality to hold true, but I don't understand why it works here. Edit: $lVert [x-1,y+1] rVert^2$ is equal to $sqrt (x-1)^2+(y+1)^2 $ analysis inequality share | cite | improve this question edited Jul 14 at 21:59 asked Jul 14 at 20:52 Direwolfox 6 2 1 What is $||(x-1),(y