I´m trying to prove that if $d$ is a metric in $X$ and $A \subset X$, then:
$$ \lvert d(x,A) - d(y,A) \rvert \le d(x,y), \phantom{5} \forall x, y \in X $$
The question seems very simple but I´m having problems to solve it. Any suggestions?
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1Check this: https://math.stackexchange.com/q/2719411/42969 or this: https://math.stackexchange.com/q/48850/42969. – Martin R Oct 02 '20 at 09:42
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1Well, just to give you an idea, first use triangular inequality for $d(x, z)$, $d(y, z)$ and $d(x, y)$ for every $z$ in $A$. Next, use the fact that $|\int f(x){\rm d} x|\leq \int |f(x)|{\rm d} x$. – ConvXET Oct 02 '20 at 09:47
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2Does this answer your question? Continuity of the function $x\mapsto d(x,A)$ on a metric space – Arnaud D. Oct 02 '20 at 09:56
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By symmetry, you just need to prove for one side that $d(x,A)-d(y,A)\leq d(x,y)$ for all $x,y\in X$. Pick $a\in A\subset X$, by triangle inequality, you get $$d(x,a)\leq d(x,y)+d(y,a)$$ Then take $\inf_{a\in A}$ on both sides, you get $$d(x,A)\leq d(x,y)+d(y,A)$$ By reordering, you get the inequality. This is a trick for this question, and I think you can figure this out why it works. Another side is similar, so I omit the proof.
Mike
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