Mapping to unique nearest point

Question

In my class, we showed that if $B$ is closed in $\mathbb{R}$ and $A\subseteq\mathbb{R}$ and every $x\in A$ has the unique nearest point $f\left(x\right)$ of $B$, then $x\mapsto f\left(x\right)$ is continuous.

My professor asked if we could find an arbitrary metric space $X$ (not necessarily with $X=\mathbb{R}$) where $A$ and $B$ (with $B$ not necessarily having to be closed now) have this same unique nearest point property, but $f$ is no longer continuous.

I've been thinking about it for a couple of days now, but I haven't come up with any examples. If anyone could provide an example, I would greatly appreciate it.

score 4 · Accepted Answer · answered Nov 22 '17 at 10:08

Notation: To avoid confusion between intervals and ordered pairs, $\langle a,b\rangle$ is an ordered pair.

With the usual metric on $\Bbb R^2,$ let $A=[0,1]\times \{0\}$ and $B= ([0,1)\times \{1\})\cup \{\langle 2,0\rangle \}. $ Let $X=A\cup B.$

If $0\leq t<1$ then $f(\langle t,0\rangle)=\langle t,1\rangle.$

But $f(\langle 1,0\rangle)=\langle 2,0 \rangle.$

So $f(x)$ is discontinuous at $x=\langle 1,0 \rangle.$ (See footnote). Note that $A$ and $B$ are closed in $X$.

The tricky part of constructing an example is that in any metric space $(X,d),$ if $X\supset B\ne \phi$ and $g(x)=\inf \{d(x,y):y\in B\}$ then $g:X\to \Bbb R$ $is$ continuous.

Footnote. The sequence $(\;f(\langle 1-1/n,0 \rangle)\;)_{n\in \Bbb N}$ does not converge to $f(\langle 1,0\rangle).$

score 1 · Answer 2 · answered Nov 22 '17 at 22:53

This is to provide an answer using an example I provided in the comments.

Let $X=\mathbb{R}^{2}$ with the standard Euclidean metric, and let $A=\left\{\left(x,0\right):x\in\mathbb{R}\right\}$ and $B=\left\{\left(x,1\right):x\ge0\right\}\cup\left\{\left(x,-1\right):x<0\right\}$. If $x\ge0$, then the only nearest point in $\mathbb{R}^{2}$ is obviously $\left(x,1\right)$. If $x<0$, then the only nearest point is obviously $\left(x,-1\right)$. Take the sequence $x_{n}=-\frac{1}{n}$. This obviously converges to $0$, but $$\lim\limits_{n\rightarrow\infty}\left(f\left(x_{n}\right)\right)=\lim\limits_{n\rightarrow\infty}\left(f\left(-\frac{1}{n}\right)\right)$$ $$\lim\limits_{n\rightarrow\infty}\left(\left(-\frac{1}{n},-1\right)\right)=\left(0,-1\right)\ne f\left(0\right)=\left(0,1\right).$$ Thus, $f$ is discontinuous at $x=0$.

score 0 · Answer 3 · edited Nov 22 '17 at 08:46

0

$A = \{1\}\times\mathbb{R};\ B = \{0\}\times\bigl([-1,0) \cup [1,2]\bigr)$.

edited Nov 22 '17 at 08:46

José Carlos Santos

427,504

answered Nov 22 '17 at 08:42

William Elliot

17,598

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What would be the unique closest point in B from (1,0.25) – Mathemagical Nov 22 '17 at 08:47
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@Mathemagical. There isn't one. – DanielWainfleet Nov 22 '17 at 08:57
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This example doesn't meet the requirements of the question then. Plus... A and B meet to be subsets of R, not R^2 – Mathemagical Nov 22 '17 at 09:01
Sorry for the confusion. $A$ and $B$ do not need to be subsets of $\mathbb{R}$. $X$ can be any arbitrary metric space. – Nov 22 '17 at 09:28
Does $A=\left{\left(x,0\right):x\in\mathbb{R}\right}$ and $B=\left{\left(x,1\right):x\ge0\right}\cup\left{\left(x,-1\right):x<0\right}$ work? – Nov 22 '17 at 09:34
@Jake yes. That works. As I understand. You mean them to be subsets of the plane. A is the x axis and B is $(\mathbf{R}^+ \cup {0}) \times {1}$ unioned with $\mathbf{R}^- \times {-1}$. – Mathemagical Nov 22 '17 at 10:02
@Mathemagical awesome. Thanks! – Nov 22 '17 at 10:05
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Since you have answered yourself, perhaps edit your question to indicate the answer you have found? – Mathemagical Nov 22 '17 at 10:07
remkved vvvvvvvvbbbnnmmm – William Elliot Nov 22 '17 at 10:18

Mapping to unique nearest point

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