If $x$ is not in the closure of a convex set $A$, is there a point in $A$ that is closer than $x$ is to each point in $A$?

Question

I'm trying to assess the following conjecture:

$\textbf{Conjecture}.$ Suppose that $A$ is a non-empty convex subset of $\mathbb{R}^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' \in A$ such that $\forall a \in A, \, ||x - a|| > || a' - a ||.$

I am familiar with results that yield this if we assume that $A$ is closed (closest point theorem, strongly separating hyperplane theorem), but I do not want to assume that $A$ is closed.

Edit: What I have in mind here is that if $A$ is closed, then there is some point $a'$ that is closest to $x$, and the perpendicular bisector of $a'x$ strongly separates $x$ and $A$, showing that $a'$ is closer to any point in $A$ than $x$ is. I'm trying to see if the conjecture holds without assuming $A$ is closed.

Answer 1

EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.

Since you referenced the closest point theorem, I thought you actually wanted to prove the following:

Suppose that $A$ is a non-empty convex subset of $\mathbb{R}^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' \in A$ such that $\|x - a'\| \le \| x - a \|$ for all $a \in A$.

This statement is wrong. Take $A=\{(x,y)\in \Bbb R^2 : x^2+y^2\le 1\}\setminus \{(0,1)\}$ and $x=(0,2)$.

Answer 2

This is a slightly heuristic argument, which I think may work. As you say there exists some $b\in \bar A$ such that $\|x-a\|>\|b-a\|$ for all $a\in A$ (where $b=P_A(x)$), to avoid triviality we assume $b\in \bar A\backslash A$. Let us first assume that $\overrightarrow{xb}$ intercepts $A$. So for all $\varepsilon>0$ there exists an $a_\varepsilon\in A\cap\overrightarrow{xb}$ such that $\|b-a_\varepsilon\|<\varepsilon$. Set $\delta=\|x-b\|/2$. Convexity implies, for all $a\in A$, that angle $\overline{ab}\angle\overline{bx}$ cannot be acute. This in combination with the fact $x,b,a_\delta$ all lie on the same line, and $\|b-a_\delta\|<\|b-x\|$ implies that $\|a-a_\delta\|<\|a-x\|$.

As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $\angle ADC<\pi/2$ we must have $|CA|<|CB|$.

If $\overrightarrow{xb}\cap A =\emptyset$, I believe we can use a similar argument, but we will have to control the value of $\delta$ more tightly (remembering that $a_\delta$ is no longer on $\overrightarrow{xb}$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.