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If we consider a bounded subset $A$ of a metric space $(X,d)$, the distance to the set $A$ is given by $$d_A(x) = \inf \{d(x,a):a\in A \}.$$ If we consider two bounded sets $A$ and $B$. I need to prove that: $$d_{A\cup B}(x) = \min(d_A(x),d_B(x) )$$ $$d_{A\cap B}(x) \geqslant \max(d_A(x),d_B(x) ).$$

I have troubles with proving it rigorously, but I can see it is true if I draw it on a paper. Can somebody help me?

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Take $d<\min\bigl(d_A(x),d_B(x)\bigr)$. If we had $d_{A\cup B}(x)=d$, then there would be a $y\in A\cup B$ with$$d<d(x,y)<\min\bigl(d_A(x),d_B(x)\bigr).$$But $y\in A$ or $y\in B$ and therefore this is impossible: if $y\in A$, then $d(x,y)<d_A(x)$ and if $y\in B$, $d(x,y)<d_B(x)$.

So $d_{A\cup B}(x)\geqslant\min\bigl(d_A(x),d_B(x)\bigr)$. Suppose we had$$d_{A\cup B}(x)>\min\bigl(d_A(x),d_B(x)\bigr).$$Then we would have $d_{A\cup B}(x)>d_A(x)$ or $d_{A\cup B}(x)>d_B(x)$. This can't be, because $A,B\subset A\cup B$ and therefore it follows from the definition of distance from a point to a set that $d_{A\cup B}(x)$ is smaller than or equal to both $d_A(x)$ and $d_B(x)$; in other words,$$d_{A\cup B}(x)\leqslant\min\bigl(d_A(x),d_B(x)\bigr).$$

By this last argument, since $A\cap B\subset A,B$, $d_{A\cap B}(x)\geqslant d_A(x),d_B(x)$. Therefore,$$d_{A\cap B}(x)\geqslant\max\bigl(d_A(x),d_B(x)\bigr).$$

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