
#1
Oct505, 12:15 PM

P: 81

Suppose f is a function from a metric space (X,D) into another metric space (Y,D') such that D(x,x') >= kD'(f(x),f(x'), where k is a constant positive real number. Prove that f is continuous.
Okay, I know that there is a theorem that says "preimages of open sets are open" so I suppose I can use that. Let U be an open set in (Y,D'). Since U is open, then there exists a neighborhood, N(f(x),p) (a pneighborhood around f(x).) such that it exists in U. By theorem, neighborhoods are always open. (so basically, I need to find a "q" radius around x such that N(x,q) is open in (X,D).) Although I know what the conclusion should be, I can't find a way to approach that solution. Could anyone give any assistance? 



#2
Oct505, 03:23 PM

HW Helper
P: 2,566

If there is some point x' in X that maps to p, then let you know p=f(x'), and you can use the assumption in the problem. The preimage of U is the set of all x that map to some p in U, and if you can show these are in an open neighborhood of x, the function is continuous. (Well technically, you should show that the preimage of any open set is open, which would use the fact that any open set is the union of open balls)



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