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Let $(X,d)$ be a connected metric space(e.g. metrizable topological vector space, or $R^n$(with $n\ge 2$)) with metric $d$, $A$ is a closed subset of $X$ with the property that for each $x∈X$, there is a unique $y∈A$ with $d(x,y)=\min_{z∈A}d(x,z)$. This defines a function $f:X→A$ by mapping all $x∈X$ to $y∈A$ as above.

Is $f:X\to A$ continuous?

Some Remarks:

(1)Of course if $f$ is continuous, then $A$ is a retract of $X$.

A very general example is that every closed convex complete subset in any inner product space has this property by Hilbert projection theorem(and in this case $f$ is a projection(Lipschitz continuous with Lipschitz constant $1$)).But I tried but failed to find a non-convex closed $A$ with this property in even $R^2$, so in $R^2$, $A$ with such property must be convex?

(2)And from this post Show that $a\to f(a)$ from $A$ to $S$ is continuous., we can see this is true for $X=\mathbb{R}$.

(3)But this $f$ may be not continuous for $X$ being only a metric space in general by a previous post Mapping to unique nearest point. (After thinking for a while, I think this counter-example is somewhat related to the un-connectedness of $X$, so I am curious about what if $X$ is connected? ).

stephenkk
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  • Just out of curiosity, can you give an example of such a space where $A$ is not all of $X$? It sounds like a pretty weird space to me. – Richard Jensen May 03 '22 at 10:44
  • I think $X = \mathbb R^2$ and $A = \mathbb R$ works, or more generally $A$ any linear subspace of a Euclidean space $X$ – SolubleFish May 03 '22 at 10:45
  • @RichardJensen This also works if $A$ is a singleton. – J.-E. Pin May 03 '22 at 10:51
  • @RichardJensen, consider a non-trivial case, let $C$ be a cantor space ${0,1}^N$ with metric $d(x,y)=\sum_{n=1,2,3,...}|x_n-y_n|/10^n$. Then we can see every closed subset of $C$ has this property. – stephenkk May 03 '22 at 11:00
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    Does this answer your question? Mapping to unique nearest point – Kavi Rama Murthy May 03 '22 at 11:37
  • Yes thanks！@KaviRamaMurthy. But I think the metric space there seems not so interesting and too special, how about letting X be a metrizable topological vector space, normed space? – stephenkk May 03 '22 at 11:41
  • You don't make sense. $\mathbb R^{2}$ a very nice topological vector space. You are not requiring $A$ to have any connection with the vector space structure (like convexity). You should spell out exactly what according to you is missing in that exmaple. – Kavi Rama Murthy May 03 '22 at 12:14
  • @KaviRamaMurthy, thanks for your comment. very much. For example in the accepted answer of this post, by construction $B=[0,1)\times{1}\cup (2,0)$, $A=[0,1]\times {0}$. This $B$ is closed in the $X=A\cup B$ in this answer. But $B$ is not closed in $R^2$(this allows the sudden discontinuity in this post I think). So I am curious about what if $X=R^2$ to avoid this. – stephenkk May 03 '22 at 12:28
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    @RichardJensen - Any closed convex set $A$ in $\Bbb R^n$ has this property. If $u, v$ are two distinct points in $A$ the same distance $r$ from some $x$, then $\frac {u+v}2$ is also in $A$, and must be closer to $x$ since $u, v$ both lie on the sphere of radius $r$ about $x$, while the line segment connecting them is inside it. – Paul Sinclair May 03 '22 at 23:26

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