Show that $d(x,A) = d(x,\bar A)$ in $\mathbb{R}^n$, $d(x,A) = \inf\{d(x,y) | y\in A\}$

Question

I try to solve this problem by the following two methods, but get stuck on both. Any wise advice ? Thanks :-)

• Since $A\subseteq \bar A$,then $d(x,A)\ge d(x,\bar A)$. And I need to show the inverse inequality.

• $\exists\; x_{0}\in \bar A$, such that $d(x,A) = d(x,x_{0})$. But how can I prove that $d(x,\bar A) = d(x,x_{0})$ ?

Answer 1

You do not need completeness. You know already $d(x,A)\ge d(x,\bar A)$.

Assume $d(x, \bar A) < d(x,A)$. Now, choose $C >0$ such that $$d(x, \bar A) < C < d(x,A) \Rightarrow \exists\,\bar x \in \bar A: \, d(x, \bar A) \leq d(x,\bar x) < C$$

Now, choose a sequence $x_n \in A$ such that $$\lim_{n\rightarrow\infty}x_n = \bar x \Rightarrow \lim_{n\rightarrow\infty}d(x,x_n) = d(x,\bar x)$$

Note that $d(x,x_n)\geq d(x,A)$. So, it follows

$$C < d(x,A)\leq d(x,\bar x) < C$$ Contradiction! $\Rightarrow d(x, \bar A) = d(x,A)$