Triangle inequality applied to sets and a point

**paupsers** · August 29th 2011, 04:16 PM

Let A and B be non-empty subsets of Y, and let $z \in Y$ . Then

$d(A,B) \leq d(A,z)+d(z,B)$ .

Obviously, this is a restatement of the triangle inequality, but I'm having a hard time putting the proof into words. Should I begin by assuming $d(A,B)=t_1=|a-b|$ , where (a,b) is in AxB? Because then I run into a problem when I assume that
$d(A,z)=t_2=|a-z|$ because I'm using the same element $a \in A$ , which I don't think I can assume.

Can anyone give me some help with this?

**vincisonfire** · August 29th 2011, 04:52 PM

Assuming that $d$ is a metric, the above is true.
Isn't the problem more like :
Here is a definition of $d$ , prove that it is a distance, and hence that it satisfies the triangle inequality.
An intuitive definition of $d$ would be $min\{d(a,b) | a\in A b\in B\}$ .
In this case you can proceed as you said from a pain $(a,b)$ such that d(a,b) = d(A,B).

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