Proving a Proposition in Metric Spaces: Finding the Intersection of Two Balls

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SUMMARY

The proposition states that in a metric space, if point c belongs to the intersection of two open balls B(a; r) and B(b; s), then there exists a radius t > 0 such that the open ball B(c; t) is entirely contained within this intersection. The proof involves using the triangle inequality and determining the distances from point c to the boundaries of the two balls. Specifically, t should be defined as the minimum distance from c to the boundaries of B(a; r) and B(b; s), ensuring that B(c; t) remains within the intersection.

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Homework Statement


I'm trying to prove this proposition:

Let a and b be points in a metric space and r, s > 0. If c belongs to the intersection of B(a; r) and B(s; b), then there exists a number t > 0 such that B(c; t) is contained in the intersection of B(a; r) and B(s; b).

(where B(a; r) = {x in M : d(x,a) < r} if (M,d) is the metric space.)

Homework Equations





The Attempt at a Solution


Let d be the metric.

Well I know what I want to prove, namely for some for t > 0, d(x,c) < t implies d(x,a) < r and d(x,b) < s.

About all I have to work withis the hypothesis that c is in the intersection of the two balls (so d(c,a) < r and d(c,b) < s) and the the triangle inequality. All I can come up is something like d(x,c) <= d(x,a) + d(a,c) < d(x,a) + r (and a similar string of inequalities for d(x,b)), which clearly doesn't do much.

I can't figure out the right t. Help please!
 
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Use your intuition to decide what t should be. Draw an open disc of radius r centered at a. Draw an open disc of radius s centered at b. Draw c in the intersection. How far is c from the boundary of the first disc? How far is c from the boundary of the second disc? Take t to be the minimum of those two distances.

Then write up a formal proof using your t.
 

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