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In showing that the diameter of a compact set $A$ is attainable, one approach is to consider a function $f:A\times A\rightarrow\mathbb{R}$ such that $f(x,y)=d(x,y)$. The key is to show that the function $f$ is continuous on $A$.

I proved the continuity by using the definition: that for $\delta=\epsilon$, if $d(a_1,b_1)+d(a_2,b_2)<\delta$, then $|d(a_1,a_2)-d(b_1,b_2)|<\epsilon$.

This follows from repeated applications of the triangle inequality: $d(a_1,a_2)-d(b_1,b_2)<(d(a_1,b_1)+d(b_1,a_2))-(d(b_1,a_2)-d(a_2,b_2)) = d(a_1,b_1)+d(a_2,b_2)<\epsilon$. Similarly, $d(b_1,b_2)-d(a_1,a_2)<\epsilon$, and hence $|d(a_1,a_2)-d(b_1,b_2)|<\epsilon$.

Are there other ways to see that $f$ is continuous? (I was hoping there's perhaps a simpler way...)

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PJ Miller
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  • is there any extra info? because you can apply your argument in a situation where you have an infinite dimensional vector space with non-continuous norm and yet have it being continuous. – Averroes Jun 18 '13 at 22:58
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    @Averroes, I believe the "discontinuous norm" is discontinuous wrt the product topology, not the metric topology. – dfeuer Jun 18 '13 at 23:15
  • @dfeuer thanks for the hint :) – Averroes Jun 18 '13 at 23:17

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