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For a point $x$ and a non-empty subset $A$ of a metric space $(X, d)$, define \begin{align}\inf\left\{ d(x,a):a\in A\right\}\end{align}

Prove that if $y$ is another point in $X$ then $$d(x,A)\leqslant d(x,y)+d(y,A)$$

The game
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    Let $\varepsilon>0$ be arbitrary.

    Then by the definition of the infimum, $d(y,A)+\varepsilon$ is not a lower bound for ${d(y,a)\mid a \in A}$ (or it would contradict the maximality of $d(y,A)$) so there is some $a_0 \in A$ so that $d(y,a_0) < d(y,A) + \varepsilon$. It follows that $$d(x,A) \le d(x,a_0) \le d(x,y) + d(y,a_0) < d(x,y) + d(y,A) + \varepsilon$$

    As this holds for arbitrary $\varepsilon >0$ we have that $d(x,A) \le d(x,y) + d(y,A)$.

     – Henno Brandsma Mar 28 '21 at 15:11

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Hint. For every $a\in A$, by triangle inequality, $$d(x,a)\leq d(x,y)+d(y,a).$$ What's next?

Bernard Pan
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