# 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 $A$?

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.

• So the problem is that $a'$ might belong to $\overline A \setminus A$, right? Mar 13, 2019 at 18:02
• @GiuseppeNegro Yes, that's exactly what I'm thinking. Mar 13, 2019 at 18:17
• Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $A\setminus\{P(x)\}$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier. Mar 13, 2019 at 20:49
• @GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning. Mar 13, 2019 at 21:37
• What about taking the closest point to $x$ in $\overline{A}$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...) Mar 13, 2019 at 22:14

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.

• Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere. Mar 14, 2019 at 3:07
• @supergeneric I hope you do. If I have time later I will try think about it again. Mar 14, 2019 at 10:28

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 $$x=(0,2)$$.

• I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit. Mar 13, 2019 at 18:12
• Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in. Mar 13, 2019 at 18:14
• No problem, thanks for looking! Mar 13, 2019 at 18:17