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I am trying with no luck to prove:

Let (X,d) be a metric space and A a non-empty subset of X. For x,y in X, prove that

d(x,A) < d(x,y) + d(y,A)

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    I've drawn all the pictures. But doesn't it come down to some creative trick or using a fact about infimum? I can't see it.2012-02-12

1 Answers 1

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If you have $\leq$ instead of $<$, then you have:

$d(x,A) = \inf_{z\in A} d(x,z)$. Now, say $z_0\in A$ and $y\in X$. Then $d(x,z_0)\leq d(x,y) + d(y, z_0)$. Taking infimum over all $z\in A$ of the left hand side, we obtain:

$ d(x, A) = \inf_{z\in A}d(x,z) \leq d(x,z_0) \leq d(x,y) + d(y, z_0). $

Observe that $d(x, A)$ is now independent of $z_0$. Hence taking the infimum over all $z$ in $A$ of the right hand side, we get:

$ d(x,A) \leq d(x,y) + \inf_{z\in A}d(y,z) = d(x,y) + d(y, A). $

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    @student: Most things in mathematics aren't easy or natural when you see them for the first time. It takes practice, no matter whether you're a student or a professional mathematician.2012-02-12