Prove that $\cup_{\alpha} A_{\alpha}$ has ﬁnite diameter

Question

Prove that $\cup_{\alpha} A_{\alpha} $ has ﬁnite diameter if $\cap_{\alpha} A_{\alpha}≠∅$ and there exists a constant $M$ such that $diam(A_{\alpha}) ≤ M$ for all $\alpha$. Each $A$ is a subset of metric space $(X,d)$, and each $A$ has a finite diameter (that is, $diam(A) = sup{d(x,y): x,y ∈A}$).

Not too sure where to begin, any help appreciated. Note ∝ is a subscript.

And btw, what have you tried? Any point in the union is at most M in distance from the intersection. — M.B., Mar 22 '14 at 23:48

score 1 · Answer 1 · answered Mar 22 '14 at 23:53

1

Pick a point in the intersection. Then every point in your union is at distance at most what of that?

answered Mar 22 '14 at 23:53

Pedro

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Prove that $\cup_{\alpha} A_{\alpha}$ has ﬁnite diameter

1 Answers1