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This proof should be easy but I get stuck because I don't know how to deal with the infimum in this case.

I want to prove that $$d(x,A)\le d(x,y)+d(y,A) $$ with $$d(x,A)=\text{inf}(\{d(x,a):a\in A\})$$

My attempt: Let $y\in X$. We have $d(x,a)\le d(x,y)+d(y,a) $ for any $a\in A$.

How can I get the inequality for the infimum from this?

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    $\begingroup$ Show that for every $\varepsilon > 0$ you have $d(x,A) \leqslant d(x,y) + \bigl(d(y,A) + \varepsilon\bigr)$. $\endgroup$ Jan 18, 2017 at 12:38

2 Answers 2

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For all $a\in A$ we have by triangle inequality

$$d(x,A)\le d(x,a)\le d(x,y)+d(y,a)$$ so $$d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$$ hence $$d(x,A)-d(x,y)\le d(y,A)$$

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  • $\begingroup$ You seem to have used $d(y,a)\le d(y,A)$... $\endgroup$
    – soap
    Jan 18, 2017 at 14:19
  • $\begingroup$ Otherwise, how can you justify the last step? Just because $d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$ does that mean that $d(x,A)-d(x,y)\le d(y,A)$? There may not exist an $a\in A$ such that $d(y,A)=d(y,a)$... $\endgroup$
    – soap
    Jan 18, 2017 at 14:25
  • $\begingroup$ @XicoSimThe answer is correct. Look at the definition of the infimum. $\endgroup$ Jan 18, 2017 at 14:46
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    $\begingroup$ @XicoSim If you have $C \leqslant d(y,a)$ for all $a\in A$, then you also have $C \leqslant \inf\limits_{a\in A} d(y,a)$. $\endgroup$ Jan 18, 2017 at 14:59
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    $\begingroup$ @XicoSim If $\inf B < D$, then in particular there is a $b\in B$ with $b < D$. So if $C \leqslant b$ for all $b\in B$, then $C \leqslant \inf B$. $\endgroup$ Jan 18, 2017 at 15:18
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Let $\epsilon>0$. Then $d(y,A)+\epsilon$ can not be a lower bound of the set $\{d(y,a):a\in A\}$. Thus, there exists $w\in A$ such that

$$d(y,w)<d(y,A)+\epsilon.$$

Thus, $$ \begin{align} d(x,A)&\leq d(x,w)\\ &\leq d(x,y)+d(y,w)\\ &<d(x,y)+d(y,A)+\epsilon. \end{align} $$ We have shown that $$d(x,A)<d(x,y)+d(y,A)+\epsilon\quad\text{for all }\epsilon>0.$$ Thus, $$d(x,A)\leq d(x,y)+d(y,A).$$

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