Is $\inf\{d(x,F):x\in E\}$ equal to $\inf\{|x-y|:x\in E,y\in F\}$?

Question

Let $E,F$ be two non-empty disjoint sets in $\mathbb{R}^n$ and let $|\cdot|$ be the Euclidean norm. As introduced in many textbooks on real analysis, we can define the distance $d(E,F)$ between $E$ and $F$ by $$d(E,F)=\inf\{|x-y|:x\in E,y\in F\}.$$ I'm wondering if we can find $d(E,F)$ by following a two-step procedure. Specifically, if we define the distance $d(x,F)$ from $x\in E$ to $F$ by $$d(x,F)=\inf\{|x-y|:y\in F\},$$ then does $$\inf\{d(x,F):x\in E\}$$ equal $d(E,F)$? I didn't come across a question regarding double infimums before, and this question right here is too abstract for me. Does anyone have an idea? Thank you.

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Update: I got something!!! Let $\epsilon>0$ be given. If we could verify the following two assertions (1) and (2), then the two infimums in the title would agree!!!

(1) $\forall x\in E$, $d(E,F)-\epsilon<d(x,F)$.

(2) $\exists x\in E$ s.t. $d(E,F)+\epsilon>d(x,F)$.

Assertion (2) can be easily verified by taking $(x,y_0)\in E\times F$ with $d(E,F)+\epsilon>|x-y_0|$ and noting that $$|x-y_0|\geq\inf\{|x-y|:y\in F\}=d(x,F).$$ Now it remains to show that assertion (1) is true.

Answer 1

Thank God! I successfully verified assertion (1)!!! Fix $x_0\in E$. Now observe that $$d(E,F)=\inf(\bigcup_{x\in E}\{|x-y|:y\in F\})\leq\inf\{|x_0-y|:y\in F\}=d(x_0,F).$$ Assertion (1) then follows immediately.

Answer 2

I’m just gonna show the assertion.
Indeed, $d(x,F)\geq d(E,F)$, since $x\in E$.
Then let $\epsilon>0$ be given. Then $d(x,F)\geq d(E,F)>d(E,F)-\epsilon$.

Answer 3

It's all the same and you don't have to go so much into the definition of distances to see that. You simply have to show that for a family $(A_i)_{i \in I}$ of subsets of $\mathbb R$ you have $$\inf_{i \in I} (\inf(A_i)) = \inf \left( \bigcup_{i \in I} A_i \right).$$