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I have a very simple question regarding proof of the distance metric d(x,A). Given non empty subset of $\mathbb R^n$ $A$ and $\forall x \in \mathbb R^n$, $ f(x) = inf\{d(x,a):a\in A\}$. Prove $f(x)$ is continuous.

Prove sited here starts of with taking two points x and y and arbitrary $a\in A$ and via triangle inequality:

(1) $d(x,a)\leqslant d(x,y)+d(y,a)$.

Then by taking infimum over a, the previous equation is:

(2) $d(x,A)\leqslant d(x,y)+d(y,A)$ and so on.

I have problem with this last step. It is not obvious to me how it is possible to go from step (1) to step (2)? Why does going from $d(x,a)$ to $d(x,A)$ and from $d(y,a)$ to $d(y,A)$ preserve inequality in (1)?

Any help to clarify this problem is hugely appreciated.

Community
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Leon
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2 Answers2

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The point to be understood here, imo, is that taking the infimum/supremum over some set does not affect anything that is not in that set.

Remember: the infimum of a set is, first of all, a lower bound of that set. Next, it is the greatest (in a very precise sense that I hope you understand) such bound.

Thus,

$$\left(\forall\,a\in A\;\forall\,x,y\in X\;,\;\;d(x,a)\le d(x,y)+d(y,a)\right)\implies $$

$$\implies d(x,A):=\inf_{a\in A}d(x,a)\le\inf_{a\in A}\left(\,d(x,y)+d(y,a)\,\right)=d(x,y)+d(y,A)$$

Timbuc
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We know that for any $a\in A$ and $x,y\in X$, $$d(x,a)\leqslant d(x,y)+d(y,a)$$

But for any $a\in A$ and $x$ fixed, $d(x,A)\leqslant d(x,a)$ so that $$d(x,A)\leqslant d(x,y)+d(y,a)$$

Since now $d(x,A)-d(x,y)$ is a lower bound on the collection of elements of the form $d(y,a)$ for $a\in A$ and $y$ fixed, and since $d(y,A)$ is the largest lower bound, it follows that $$d(x,A)-d(x,y)\leqslant d(y,A)$$ obtaining the desired inequality.

Pedro
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