f(x) is in R because f(x) is a distance, whowever I think you nedd to prove d(xn,a)->d(x,a)
Fix a point a in a metric space (M,d), and define f:M->R by f(x)=d(x,a). Show f is continuous on M.
I took a sequence Xn->x in (M,d)
So f(Xn)=d(Xn,a)
taking the limit, f(Xn)->d(x,a)=f(x)
Since we have Xn->x => f(Xn)->f(x), f is continuous on M.
But I feel like that's wrong, because how do we know f(x) is in R?
I need to show if Xn->x in (M,d), then f(Xn)->f(x) in R.....but I don't know whether or not f(x) is in R do I? Am i missing something obvious?
Thanks guys.
If is a function (where are metric spaces) then is said to be Lipschitz with Lipschitz constant if . I actually didn't prove Lipschitz per se (since I fixed ) but it can easily be proved using the EXACT same idea that is Lipschitz.
Every Lipschitz function is uniformly continuous and thus continuous. It's clear for if is given then choosing we see that