Proving Continuity in Metric Spaces Using Open Sets

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The discussion centers on proving the continuity of a function f from a metric space (X,D) to another metric space (Y,D') under the condition that D(x,x') ≥ kD'(f(x),f(x')), where k is a positive constant. The key theorem referenced is that pre-images of open sets are open, which is essential for establishing continuity. The approach involves demonstrating that for any open set U in (Y,D'), the pre-image of U contains an open neighborhood around x in (X,D). This leads to the conclusion that if the pre-image of every open set is open, then f is continuous.

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Mathematicians, students of analysis, and anyone studying topology or metric spaces will benefit from this discussion, particularly those interested in the formal proofs of continuity in mathematical functions.

TimNguyen
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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 "pre-images 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 p-neighborhood 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?
 
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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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