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This article is cited in 3 scientific papers (total in 3 papers) Properties of at most countable unions of pairwise disjoint sets in asymmetric spaces I. G. Tsar'kovab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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    § 1. IntroductionThe present paper considers several important problems in approximation theory in normed linear spaces and more general linear spaces equipped with an asymmetric norm or seminorm $\|\,{\cdot}\,|$. An asymmetric norm $\|\,{\cdot}\,|$ on a real linear space $X$ is defined by the axioms: (1) $\|\alpha x|=\alpha\| x|$ for all $\alpha\geqslant 0$, $x\in X$; (2) $\| x+y|\leqslant \| x |+\| y|$ for all $x,y\in X$ and (3) $\|x|\geqslant 0$ for each $x\in X$, and (3a) $\|x|= 0$ $\Leftrightarrow $ $x=0$. Any asymmetric norm is generated by the Minkowski functional of some (asymmetric, in general) body containing the origin in its kernel. For any asymmetric norm $\|\,{\cdot}\,|$ we construct its symmetrization norm defined by $\|x\|_{\mathrm{sym}}:=\max\{\|x|,\|-x|\}$, $x\in X$. The ‘closed’ and open balls with centre $x$ and radius $r$ in a space $X=(X,\|\,{\cdot}\,|)$ with asymmetric (semi)norm are defined, respectively, by
$$
\begin{equation*}
B(x,r):=\{y\in X\mid \|y-x|\leqslant r\} \quad\text{and}\quad \mathring{B}(x,r):=\{y\in X\mid \|y-x|< r\}.
\end{equation*}
\notag
$$
The ball $B(x, r)$ can fail to be closed in the topology generated by the subbase $\tau$ of open balls. Note that if the unit ball $B(0,1)$ is closed in $X$, then so is each ball $B(x,r)$. In this case the asymmetric space $X$ is Hausdorff, and the asymmetric norm is continuous in the topology $\tau$ (see [1]). Below, unless otherwise specified, $X$ will be equipped with the default topology $\tau$, and so open and closed sets in $X$ are defined relative to $\tau$. Likewise, a mapping $f\colon E\to X$ from a topological space $(E,\upsilon)$ into $X$ is continuous if it is continuous from $(E,\upsilon)$ to $(X,\tau)$. The sphere with centre $x$ and radius $r$ is defined, as usual, by For brevity, we set $S=S(0,1)$. We also define
$$
\begin{equation*}
B^-(x,r):=\{y\in X\mid \|x- y|\leqslant r\} \quad\text{and}\quad \mathring{B}^-(x,r):=\{y\in X\mid \|x- y|< r\}.
\end{equation*}
\notag
$$
The topology $\tau$ generated by the subbase of open balls $\{\mathring{B}(x,r)\}_{x,r}$ will be also referred to as the left topology; the right topology $\tau^-$ is generated by the balls $\{\mathring{B}^-(x,r)\}_{x,r}$. In general, a space with an asymmetric norm (an asymmetric space) satisfies only the $T_1$ separation axiom (that is, for all $a, b\in X$ one can find their neighbourhoods $O(a)$ and $O(b)$ such that $a\notin O(b)$ and $b\notin O(a)$), and can fail to be Hausdorff. We will also consider asymmetric seminorms $\|\,{\cdot}\,|$, which are defined by the above axioms (1)–(3) with (3a) replaced by the condition $(\|x|= 0=\|-x|)$ $\Rightarrow$ $x=0$. An asymmetric space $(X,\|\,{\cdot}\,|)$ equipped with an asymmetric seminorm will simply be called an (asymmetric) seminormed space. Any asymmetric seminormed space satisfies the $T_0$ separation axiom. The case of seminormed spaces will always be explicitly mentioned; otherwise a space will be assumed to be equipped with an asymmetric norm. An asymmetric (semi)norm $\|\,{\cdot}\,|$ is upper semicontinuous on $X$ (see [2], Proposition 1.1.8); this follows from the fact that the set $\{x\in X\mid \|x|\in (-\infty,a)\}=\mathring{B}(0,a)$ is open in the topology generated by the subbase of open balls. Given an asymmetric (semi)norm $\|\,{\cdot}\,|$ on $X$, the symmetrization (semi)norm is defined by
$$
\begin{equation*}
\|x\|=\|x\|_{\mathrm{sym}}:=\max\{\|x|,\|-x|\}, \qquad x\in X.
\end{equation*}
\notag
$$
In a finite-dimensional space any asymmetric norm is equivalent to any norm on this space and, in particular, to the symmetrization norm. The symmetric norm $\|\,{\cdot}\,\|_{\Sigma}$ defined by is also useful. Note that $(X,\|\,{\cdot}\,|)$ is Hausdorff if and only if $\|x\|_\Sigma$ is a norm (see [3]). It was also shown in [3] that $\|\,{\cdot}\,\|_{\Sigma}$ is the supremum of the symmetric seminorms $\|\,{\cdot}\,\|$ satisfying $\|\,{\cdot}\,\| \leqslant\|\,{\cdot}\,|$, $\|-{\cdot}\,|$.Consider the homogeneous symmetric nonnegative functional measuring the length of a line segment $[x, y]$ (the ‘minimum length’ functional).As already noted, $\|\,{\cdot}\,\|_{\Sigma}\leqslant\|\,{\cdot}\,|$, $\|-{\cdot}\,|$, and so $\|x-y\|_\Sigma\leqslant \ell(x,y)$; in addition, the unit ball with respect to the seminorm $\|\,{\cdot}\,\|_\Sigma$ coincides with the closure of the convex hull of $B(0,1)\cup B^-(0,1)$. Given nonempty sets $A,B\subset X$, we set Asymmetric norms (the Minkowski functionals) seem to date back to Minkowski (see [4]). M. G. Krein was the first to employ asymmetric norms in infinite-dimensional analysis (see [5]); he also proposed the term ‘asymmetric norm’ in 1938. Krein used asymmetric norms to deal with the Markov moment problem and associated extremal problems. The significance of sublinear functionals (another name for asymmetric seminorms) for some problems in convex analysis and calculus goes back to König [6], [7]. Asymmetric distances are also natural in approximation theory (for example, in problems of one-sided approximation); Babenko [8], for example, pointed out that asymmetric approximations can be looked upon as a ‘bridge’ between problems of best approximation and best one-sided approximation (see Dolzhenko and Sevast’yanov [9], [10], and also Collatz and Krabs [11], Ch. I, § 9.E, and Cobzaş [2] in this regard). Asymmetric norms have also proved to be useful in complexity analysis in computer science (see [12]). For problems and results on asymmetric distances in geometric approximation theory and convex analysis, see Borodin [13], Ivanov [14], Ivanov and Lopushanski [15], [16], Alimov [17]–[19] and this author, on his own [20]–[27] and in collaboration with Alimov [28]–[30]. In this paper we study the classical concept of approximative compactness, and show that Hausdorff asymmetric normed spaces are precisely asymmetric normed spaces where each approximatively compact set is closed (Corollary 1). We also characterize the asymmetric normed spaces where each nonempty approximatively compact set (or nonempty compact set) is proximinal (Theorem 2). A sufficient condition for a nonempty compact subset of an asymmetric normed linear space to be compact is also provided (Theorem 4). Next, we solve one long-standing problem in normed linear spaces in the more general setting of asymmetric normed spaces. Namely, Klee [31] constructed an example of a discrete Chebyshev subset of a Banach space $\ell^1(\omega)$, where $\omega$ is a regular infinite cardinal such that $\omega^{\aleph_0}=\omega$. The Chebyshev set in Klee’s example is a disjoint union of uncountably many points. This suggests the problem of whether a normed linear space can contain an at most countable nonsingleton Chebyshev set. In our paper we give a negative answer to this question by showing that, even in not necessarily complete asymmetric normed linear spaces an at most countable nonsingleton set is never a Chebyshev set (Corollary 2). We prove that an at most countable union of pairwise disjoint proximinal sets $\{M_j\}_{j\in J}$, $\operatorname{card}J\geqslant 2$, in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ is not $P$-path-connected (Theorem 6). We also show that, in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ where the unit ball $B(0,1)$ is closed, an at most countable approximatively compact union of pairwise disjoint closed sets $\{M_j\}_{j\in J}$, $\operatorname{card}J\geqslant 2$, is not $P$-connected (Corollary 7). § 2. Properties of compact and approximatively compact sets in asymmetric spacesDefinition 1. A subset $M$ of an asymmetric normed linear space $X=(X,\|\,{\cdot}\,|)$ is compact if any open cover of $M$ contains a finite subcover. A set $M\subset X$ is boundedly compact if its intersection with any ball $B(x,r)$ is compact. A set $M\subset X$ is regularly boundedly compact if its intersection with any closed bounded set is compact. Remark 1. Any compact subset of an asymmetric normed linear space $X$ is a regularly boundedly compact set. If the unit ball $B(0,1)$ of $X$ is closed, then so is each ball $B(x,r)$, and therefore $M$ is boundedly compact in $X$ if and only if it is regularly boundedly compact. Given $x\in X$ and a nonempty $M\subset X$, the distance from $x$ to $M$ is defined by Definition 2. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space, and let ${M\subset X}$. We say that $x\in X$ is a point of approximative compactness for $M$ (written $x\in \operatorname{AC}(M)$) if any minimizing sequence $\{y_n\}\subset M$ from $M$ for $x$ (which means that $\|y_n-x|\to\varrho(x,M)$ as $n\to\infty$) contains a subsequence converging to some point $y\in M$. If $\operatorname{AC}(M)=X$, then we say that $M$ is approximatively compact. Remark 2. Any nonempty compact subset of an asymmetric normed linear space is approximatively compact. Also note that if a sequence $\{y_n\}$ converges to some point $y$, then the set $M:=\{y_n\mid n\in \mathbb{N}\}\cup \{y\}$ is compact and approximatively compact. Indeed, any sequence contained in a compact set has a subsequence converging to some point of this set. Of course, this is also true for any minimizing sequence of points in this compact set for any point of the space. Hence any nonempty compact set is approximatively compact. Let us show that the set $M:=\{y_n\mid n\in \mathbb{N}\}\cup \{y\}$ is compact. Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $M$. Then there exists $\alpha_0\in A$ such that $y\in G_{\alpha_0}$. The complement of $G_{\alpha_0}$ contains only a finite number of terms of the convergent sequence $\{y_n\}$. The remaining terms in $\{y_n\}$ are covered by the finite set $\{G_{\alpha_j}\}_{j=1}^N$. Hence $\{G_{\alpha_j}\}_{j=0}^N$ is a finite subcover of $M$. Hence $M$ is compact and approximatively compact. Theorem 1. Any non-Hausdorff asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ contains a compact (an approximatively compact) set $M\subset X$ such that $x=0\notin M$ and $\varrho(x,M)=0$. Proof. There exist two different points $x$ and $y$ such that $\mathring{B}(x,\varepsilon)\cap \mathring{B}(y,\varepsilon)\neq\varnothing$ for each $\varepsilon>0$. Translating if necessary, we can assume without loss of generality that $x=0$. For $n\in \mathbb{N}$, let $z_n$ be an arbitrary point in $\mathring{B}(x,1/n)\cap \mathring{B}(y,1/n)$. By construction the sequence $\{z_n\}$ converges to $y$ and to $x=0$. There exists $\varepsilon>0$ such that $y\notin O_\varepsilon(x)$, and so we may assume that $z_n\neq x$ for each $n\in \mathbb{N}$. It is easily seen that is a compact set, $x=0\in \operatorname{AC}(M)$, and $\varrho(x,M)=0$. Indeed, for each $z\in X$, each minimizing sequence from $M$ for $z$ is, starting with some index, a subsequence of the sequence $\{z_n\}_{n=0}^{\infty}$, $z_0:=y$, and therefore this sequence converges to $y\in M$. Hence $z\in \operatorname{AC}(M)$. Since $z$ is arbitrary, the set $M$ is approximatively compact. Finally, since $\|z_n-x|\to 0$ as $n\to\infty$, we have $\varrho(x,M)=0$, which proves Theorem 1. Note that the set $M$ constructed in Theorem 1 is not closed. Corollary 1. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space. Then any approximatively compact set in $X$ is closed if and only if $(X,\|\,{\cdot}\,|)$ is Hausdorff. Proof. Necessity. Assume on the contrary that $(X,\|\,{\cdot}\,|)$ is non-Hausdorff. The set $M$ constructed in Theorem 1 is compact and approximatively compact, but $x=0$, which is a limit point of $M$, does not lie in $M$. Hence $M$ is not closed.
Sufficiency. Let $x$ be a limit point of an arbitrary approximatively compact set $M$ in a Hausdorff space $(X,\|\,{\cdot}\,|)$. There exists a sequence $\{x_n\}\subset M$ such that $x_n\to x$ as $n\to\infty$. Since $M$ is approximatively compact, there exists a subsequence $\{x_{n_k}\}$ converging to some $y\in M$. Since $(X,\|\,{\cdot}\,|)$ is Hausdorff and $\{x_{n_k}\}$ converges to $x$, it follows that $x=y\in M$. Hence $M$ is closed. This proves Corollary 1. The set of all best approximants (nearest points) from $M$ for $x$ is defined by The mapping $P_M$ is called the metric projection operator onto the set $M$.A set $M$ is a Chebyshev set in $X$ if each point in $X$ has a unique nearest point in $M$. Lemma 1. Any asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ with nonclosed unit ball $B(0,1)$ contains a nonempty compact approximatively compact set $M$ such that no minimizing sequence for $x=0$ from $M$ converges to a point in $P_M(x)$. In addition, $P_M(x)=\varnothing$. Proof. Let $(X,\|\,{\cdot}\,|)$ be a non-Hausdorff asymmetric normed linear space. By Theorem 1 there exists a compact approximatively compact set $M\subset X$ such that $x=0\in \operatorname{AC}(M)\cap \overline{M}$, $x\notin M$, and each minimizing sequence for $x$ from $M$ converges to some point $y\in M$, $y\neq x$. For each point $z\in M$ we have $0=\varrho(x,M)<\|z-x|$, that is, $y,z\notin P_M(x)$. Hence $P_M(x)=\varnothing$.
Assume now that $(X,\|\,{\cdot}\,|)$ is a Hausdorff asymmetric normed linear space with nonclosed unit ball $B(0,1)$. There exists a sequence $\{s_n\}\subset B(0,1)$ converging to some point $s$ such that $\|s|>1$. We claim that $\|s_n|\nrightarrow 0$, $n\to\infty$. In fact, if $\|s_n|\to 0$ as $n\to\infty$, then $\{s_n\}$ would converge both to 0 and to $s$, which is impossible in Hausdorff spaces. There exists a subsequence $\{t_k:=s_{n_k}\}$ such that $\|s_{n_k}|\to r\in (0,1]$ as $k\to\infty$. We have
$$
\begin{equation*}
\begin{gathered} \, \biggl\|\frac{t_k}{\|t_k|}-\frac{s}{r}\biggr|\leqslant \biggl\|\frac{t_k}{\|t_k|}-\frac{s}{\|t_k|}\biggr| +\biggl\|\frac{s}{\|t_k|}-\frac{s}{r}\biggr| =\frac{ \| {t_k}-s |}{\|t_k|}+\|s\|_{\mathrm{sym}}\biggl|\frac{\|t_k|-r}{r\|t_k|}\biggr| \to 0, \\ k\to\infty. \end{gathered}
\end{equation*}
\notag
$$
Thus, the sequence $\{y_k:={t_k}/{\|t_k|}\}\subset S$ converges to the point $y:=s/r$, $\|y|>1$. Consider the set
$$
\begin{equation*}
M:=\biggl\{z_n:=\biggl(\frac{n+1}{n}\biggr)y_n\biggm| n\in \mathbb{N}\biggr\}\cup\{y\}.
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
\|z_n-y|\leqslant \biggl\|z_n -\frac{n+1}{n}y\biggr|+\frac{1}{n}\|y| =\frac{n+1}{n}\|y_n-y|+\frac{1}{n}\|y|\to 0, \qquad n\to\infty,
\end{equation*}
\notag
$$
and therefore the set $M$ is compact and approximatively compact. Further, we have $1<\|z_n-0|=((n+1)/{n})\|y_n|\to 1$ as $n\to\infty$, and so $P_M(0)=\varnothing$, which proves Lemma 1. Remark 3. Let $(X,\|\,{\cdot}\,|)$ be a Hausdorff asymmetric normed linear space with nonclosed unit ball $B(0,1)$. Let $M$ be the set constructed in Lemma 1 in this setting. Then there exists $\varepsilon>0$ such that the ball $B(x,1+\varepsilon)$ does not contain the point $y\in M$ (see the proof of Lemma 1), but $B(x,1+\varepsilon)$ contains the elements of the sequence $\{z_n\}$ starting with some index. Hence $M\cap B(x,1+\varepsilon)$ is not compact, and so $M$ is not boundedly compact. Lemma 2. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space where the unit ball $B(0,1)$ is closed, $\varnothing\ne M\subset X$ and $x\in \operatorname{AC}(M)$. Then each minimizing sequence $\{y_n\}$ for $x$ from $M$ contains a subsequence converging to some point in $P_M(x)$. In addition, $P_M(x)$ is a nonempty compact set. Proof. We have $x\in \operatorname{AC}(M)$, and so each minimizing sequence $\{y_n\}$ for $x$ from $M$ contains a subsequence $\{y_{n_k}\}$ converging to some point $y\in M$. By assumption the ball $B(x,\varrho(x,M)+\varepsilon)$ is closed, and so $B(x,\varrho(x,M)+\varepsilon)$ contains the elements of this subsequence starting with some index. So the limit point of this sequence lies in this ball. Thus, $y\in M\cap B(x,\varrho(x,M)+\varepsilon)$ for each $\varepsilon>0$. As a result, $\|y-x|=\varrho(x,M)$, that is, $y\in P_M(x)$, proving Lemma 2. Lemma 3. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space where the unit ball $B(0,1)$ is closed, and let $M$ be a nonempty boundedly compact subset of $X$. Then $M$ is approximatively compact and $P_M(x)$ is nonempty for each $x\in X$. Proof. Let $x\in X$ and $\{y_n\}\subset M$ be such that $\|y_n-x|\to\varrho(x,M) $ as $n\to\infty$. We set $R:=\sup_{n}\|y_n-x|$. The sequence $\{y_n\}$ is contained in the compact set $M\cap B(x,R)$, and therefore it has a subsequence $\{y_{n_k}\}$ converging to some point $y\in M\cap B(x,R)$. Hence $x\in \operatorname{AC}(M)$, and the set $P_M(x)$ is nonempty by Lemma 2. This proves Lemma 3. Theorem 2. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space. Then the following conditions are equivalent: (1) the unit ball $B(0,1)$ of $X$ is closed; (2) each nonempty approximatively compact set in $X$ is proximinal; (3) each nonempty compact (regularly boundedly compact) set in $X$ is proximinal. Proof. The equivalence of assertions (1) and (2) is secured by Lemmas 1 and 2. Now the equivalence (2)$\,\Leftrightarrow\,$(3) follows from Lemmas 1–3. Definition 3. Let $U\subset X$ be a nonempty subset of an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Given $a\in U$ and $b\in X\setminus U$, the relative boundary points of the set $[a,b]\cap U$ in the line segment $[a,b]$ are called surface points of the set $U$. The set of all surface points of $U$ is called the surface of $U$ (denoted $S_U$). Note that if $X\setminus U\neq\varnothing$, then for each $b\in X\setminus U$ the line segment $[a,b]$ has a nonempty intersection with $S_U$. If $U$ is an open ball $\mathring{B}(x,r)$, $r>0$, then $S_U=S(0,1)$. Theorem 3. Let $U\subset X$ be a nonempty open subset of an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $U\cup S_U$ is closed if and only if so is the surface $S_U$. Proof. The case $X=U$ is trivial because $S_U=\varnothing$. Assume that $X\neq U$.
Necessity. If $U\cup S_U$ is closed, then so is $S_U=(U\cup S_U)\setminus U$. Sufficiency. Assume that the surface $S_U$ is closed. Consider an arbitrary convergent sequence $\{x_n\}\subset U\cup S_U$. Let $x$ be a limit point of this sequence. If $x\notin U\cup S_U$, then, since $S_U$ is closed, we can assume (by passing to a subsequence if necessary) that $\{x_n\}\subset U$. So $[x_n,x]$ contains a point $y_n\in S_U$, $n\in \mathbb{N}$. We have $\|y_n-x|\leqslant \|x_n-x|\to 0$ as $n\to\infty$, and as $S_U$ is closed, we obtain $x\in S_U$, which is a contradiction. This proves Theorem 3. Remark 4. The asymmetric norm $\|\,{\cdot}\,|$ is upper semicontinuous, and therefore the sphere $S(0,1)$ is closed if and only if $\overline{S(0,1)}\subset B(0,1)$. Remark 5. The proof of the sufficiency part in Theorem 3 shows that if $S_U$ is closed, then $U\cup S_U$ is too, without the assumption that $U$ is open. Moreover, the closedness of $U\cup S_U$ follows from the condition $\overline{S_U}\subset U$. Lemma 4. Any asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ with nonclosed unit ball $B(0,1)$ contains a compact set $M$ which is not boundedly compact. Proof. By Theorem 3 the sphere $S=S_{B(0,1)}$ is not closed, and therefore there exists a sequence $\{y_n\}\subset S$ that has a limit point $y$ outside $S$. It is known that the asymmetric norm $\|\,{\cdot}\,|$ is upper semicontinuous on $X$, and so $\|y|>1$. We set
$$
\begin{equation*}
\alpha_n:=\frac{n(\|y_n\|_{\mathrm{sym}})}{1+n(\|y_n\|_{\mathrm{sym}})}
\end{equation*}
\notag
$$
and let $z_n:=\alpha_ny_n$, $n\in \mathbb{N}$. We have $1-\alpha_n=1/(1+n(\|y_n\|_{\mathrm{sym}})$, $\|z_n|=\alpha_n\|y_n|=\alpha_n\to 1$, and since $(1-\alpha_n)\|-y_n|\leqslant (1-\alpha_n)\|y_n\|_{\mathrm{sym}}\to 0$ as $n\to\infty$, we have
$$
\begin{equation*}
\|z_n-y|\leqslant \|y_n-y|+\|z_n-y_n|=\|y_n-y|+(1-\alpha_n)\|-y_n|\to 0, \qquad n\to\infty.
\end{equation*}
\notag
$$
The set is compact, and $\|z_m|<1$ for all $m\in \mathbb{N}$. As already noted, the asymmetric norm is upper semicontinuous on $X$, and therefore no subsequence $\{z_n\}$ can converge to any $z_m$ because the norms of its elements tend to $1$. Hence for each $m\in \mathbb{N}$ there exists $\delta_m>0$ such that the neighbourhood $O_{\delta_m}(z_m)$ contains only a finite number of elements of the sequence $\{z_n\}$. Therefore, the open cover $\{O_{\delta_n}(z_n)\}$ of $M_0:=M\setminus \{y\}$ contains no finite subcover. Hence $M_0$, and therefore also $M\cap B(0,1)=M_0$, are not compact. As a result, $M$ is not boundedly compact. This proves Lemma 4. Theorem 4. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric normed linear space. Then the unit ball $B(0,1)$ is closed if and only if each compact subset of $X$ is boundedly compact. Proof. If the ball $B(0,1)$ is closed, then so is each ball $B(x,r)$, and so the intersection of any compact subset $M$ of $X$ with $B(x,r)$ is a closed subset of $M$, which therefore is a compact set. If the ball $B(0,1)$ is nonclosed, then by Lemma 4 the space $X$ contains a compact set that is not boundedly compact.
This proves Theorem 4. § 3. Properties of the metric projection to an at most countable unions of pairwise disjoint sets in asymmetric spacesIn this section we consider the problem of whether discrete (in a certain sense) sets can be Chebyshev sets, or their metric projection operators can have connected values. Theorem 5. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint proximinal sets $M_j$ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $M$ is not a Chebyshev set. Proof. Assume for a contradiction that $M$ is a Chebyshev set. We have $J=\mathbb{N}$ or $J=\{ 1,\dots,N\}$, $N\geqslant 2$. Consider arbitrary points $a\in M_1$ and $b\in M_2$. By assumption $a\neq b$. The restriction of the distance function $\varrho(\,{\cdot}\,,M_j)$, $j\in J$, to $[a,b]$ is continuous on $[a,b]$, because the restrictions of the asymmetric norm and the symmetrization norm $\|\,{\cdot}\,\|_{\mathrm{sym}}:=\max\{\|\,{\cdot}\,|,\|-{\cdot}\,|\}$ to this line segment are equivalent. For each $j\in J$ the set is closed. Indeed, consider a point $x\in [a,b]$ and a sequence $\{x_n\}\subset K_j$ such that $x_n\to x$ as $n\to\infty$. Then we have The set $M_j$ is proximinal, and therefore $\varnothing\neq P_{M_j}(x)\subset P_{M}(x)$. Since $M$ is a Chebyshev set, we have $P_{M}(x)=P_{M_j}(x)$. As a result, $x\in K_j$. Hence each $K_j$, $j\in J$, is a closed subset of $[a,b]$. The cases of a finite $J$ or a finite number of nonempty sets $K_j$ in $[a,b]$ are impossible, because this would contradict the connectedness of $[a,b]$. So we can assume without loss of generality that $\{K_j\}_{j\in \mathbb{N}}$ consists of pairwise disjoint nonempty compact subsets of $[a,b]$ that cover this segment.
The boundary points of each compact set $\{K_j\}_{j\in \mathbb{N}}$ in this family (and, in particular, of $K_1$ and $K_2$) lie on $[a,b]$. In addition, any neighbourhood of such a boundary point (in $(a,b)$) of an arbitrary compact set $K_j$ must contain boundary points of infinitely many compact sets from the family $\{K_j\}_{j\in \mathbb{N}}$. This follows from the fact that if a nonempty interval $I$ meets some closed set $K\not\supset I$ in this family (note that $K\cap K_j=\varnothing$), then this interval cannot be covered by the interior and complement of $K$ because $I$ is connected. So this set $K$ must have a boundary point in $I$. By connectedness, no neighbourhood of a boundary point of a compact set $K_j$ can be covered by a finite number of pairwise disjoint compact sets from our family, and so there are infinitely many compact sets in our family that have a boundary point in this neighbourhood. We set $\{\varepsilon_k=2^{-k}\}$ and $\{n_k\}\uparrow\uparrow +\infty$ and perform induction on $k$. 1. Let $x_1\in \partial K_1$, $x_1\neq a$ (in this case $x_1\neq b$). The neighbourhood $O_{\varepsilon_1}(x_1)$ has a nonempty intersection with infinitely many compact sets from $\{K_j\}_{j\in \mathbb{N}}$. Let $\delta_1\in (0,\varepsilon_1)$ be such that any compact set $K_j$ which meets the ball $B(x_1,\delta_1)$, except $K_1$, has an index $j>\max\{n_1,1\}$ (here we can proceed as in the previous paragraph, because $B(x_1,\delta_1)$ meets infinitely many compact sets $K_j$). As in the previous paragraph, there exist a compact set $K_m$, $m>n_1$, and a point $x_2\in \partial K_m\cap O_{\delta_1}(x_1)$. 2. Assume that we have constructed numbers $\delta_m\in (0,\varepsilon_m)$, $m=1,\dots,k-1$, and points $x_m$, $m=1,\dots,k$, such that for some compact sets $K_{j_m}$, $m=1,\dots,k$, where $j_{m+1}>\max\{j_m,n_{m+1}\}$, $m=1,\dots,k- 1$, we have $x_{m+1}\in \partial K_{j_{m+1}}\cap O_{\delta_m}(x_m)$ and $O_{\delta_{m+1}}(x_{m+1})\subset O_{\delta_m}(x_m)$, $m=1,\dots,k-2$, $x_{j_1}=x_1$. Any neighbourhood $O_{\varepsilon}(x_k)$ meets infinitely many compact sets from the family $\{K_j\}_{j\in \mathbb{N}}$. Let $\delta_{k}\in (0,\varepsilon_{k})$ be such that each compact set, except $K_{j_k}$, that intersects the ball $B(x_k,\delta_{k})$ has an index exceeding both $n_{k+1}$ and $j_k$; we also have $O_{\delta_{k}}(x_k)\subset O_{\delta_{k-1}}(x_{k-1})$. We choose some compact set $K_{j_{k+1}}$, $j_{k+1}>\max\{j_k,n_{k+1}\}$ and some boundary point $x_{k+1}$ of this set so that $x_{k+1}\in K_{j_{k+1}}\cap O_{\delta_{k}}(x_k)$. 3. Using induction we construct a sequence of points $\{x_k\}\subset [a,b]$ which is a Cauchy sequence such that $|x_{k+1}-x_k|<2^{-k}$ for $k\in \mathbb{N}$. By construction the limit point $x_0$ of this sequence is different from $a$ and $b$, and therefore it lies in $(a,b)$. In addition, $x_0\in \bigcap_{k=1}^{\infty}B(x_k,\delta_k)$, and so, by construction, $x_0$ does not lie in any compact set in the family $\{K_j\}_{j\in \mathbb{N}}$, which contradicts the condition $[a,b]\subset \bigcup_{j=1}^{\infty}K_j$. This proves Theorem 5. Corollary 2. An at most countable nonsingleton subset of an asymmetric linear space is never a Chebyshev set. This claim follows from Theorem 5 since any one-point set is proximinal. The following result is a consequence of Theorems 2 and 5. Corollary 3. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint approximatively compact sets $M_J$ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ whose unit ball $B(0,1)$ is closed. Then $M$ is not a Chebyshev set. Definition 4. A subset $M\subset X$ of an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ is $P$-path connected (respectively, $\mathring{B}$-path connected) if $P_M(x)$ is path connected for each $x\in X$ (respectively, if the intersection of $M$ with arbitrary ball $\mathring{B}(x,r)$ is path connected). Lemma 5. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint proximinal sets $M_j $ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Assume that for some $x\in X$ and $j\in J$ the set $P_M(x)$ is path connected and $P_M(x)\cap M_j\neq \varnothing$. Then $P_M(x)\subset M_j$. Proof. Assume for a contradiction that $P_M(x)\cap M_i\neq \varnothing$ for some $M_i$, $i\neq j$. In this case there exist points $A\in P_M(x)\cap M_j$ and $B\in P_M(x)\cap M_i$ and a continuous function $\varphi\in C([a,b],P_M(x))$ such that $\varphi(a)=A$, $\varphi(b)=B$. The sets are closed and pairwise disjoint; in addition, the family $\{K_m\}_{m\in J}$ covers $[a,b]$. Proceeding as in the proof of Theorem 5, we arrive at a contradiction. So we have $P_M(x)\subset M_j$, which proves the lemma. Theorem 6. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint proximinal sets $M_j$ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $M$ is not $P$-path connected. Proof. As in the proof of Theorem 5, consider arbitrary points $a\in M_1$ and $b\in M_2$. By assumption $a\neq b$. The restriction of any distance function $\varrho(\,{\cdot}\,,M_j)$, $j\in J$, to $[a,b]$ is continuous on $[a,b]$, because the restrictions of the asymmetric norm and symmetrization norm to this line segment are equivalent. For each $j\in J$ the set
$$
\begin{equation*}
K_j:=\{x\in [a,b]\mid P_M(x)\subset M_j\}=\{x\in [a,b]\mid P_M(x)\cap M_j\neq \varnothing\}
\end{equation*}
\notag
$$
is closed. Indeed, let $x\in [a,b]$, and let $\{x_n\}\subset K_j$ be such that $x_n\to x$ as $n\to\infty$. Then
$$
\begin{equation*}
\varrho(x,M)=\lim_{n\to\infty}\varrho(x_n,M)=\lim_{n\to\infty}\varrho(x_n,M_j)=\varrho(x,M_j).
\end{equation*}
\notag
$$
Since $M_j$ is proximinal, we have $\varnothing\neq P_{M_j}(x)\subset P_{M}(x)$, and therefore $P_{M}(x)=P_{M_j}(x)$ by Lemma 5. As a result, $x\in K_j$. Hence the $K_j$, $j\in J$, are closed pairwise disjoint subsets of $[a,b]$. Now the rest of the proof proceeds as in Theorem 5.
This proves Theorem 6. The following result is now a consequence of Theorems 6 and 2. Corollary 4. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint approximatively compact sets $M_j$ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ whose unit ball $B(0,1)$ is closed. Then $M$ is not $P$-path connected. Theorem 7. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint closed sets $M_j$ in an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $M$ is not $\mathring{B}$-path connected. Proof. Assume for a contradiction that $M$ is a $\mathring{B}$-path-connected set. Consider a ball $\mathring{B}(x,r)$ that intersects some $M_{j}$ and $M_{i}$, $i\neq j$ at points $A$ and $B$, respectively. Then there exists a continuous function $\varphi\in C([a,b],M\cap \mathring{B}(x,r))$ such that ${\varphi(a)=A}$ and $\varphi(b)=B$. Consider the sets The rest of the proof proceeds as in Lemma 5. This proves Theorem 7. From Theorem 7 and Corollary 1 we obtain the following result. Corollary 5. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an at most countable union of pairwise disjoint approximatively compact sets $M_j$ in a Hausdorff asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $M$ is not $\mathring{B}$-path connected. Definition 5. A sequence $\{x_n\} \subset X$ is a Cauchy sequence if for each $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that $\|x_m-x_n|<\varepsilon$ for all $m\geqslant n\geqslant N$. An asymmetric space $X=(X,\|\,{\cdot}\,|)$ is said to be right complete (respectively, left complete) if for each Cauchy sequence $\{x_n\}\subset X$ there exists a point $x\in X$ such that $\|x-x_n|\to 0$ (respectively, $\|x_n-x|\to 0$) as $n\to\infty$. A right-complete space will simply be called a complete space. Definition 6. We say that a set $M$ is not $\ell_0$-connected if there exist nonempty sets $M_1,M_2\subset X$, $M=M_1\sqcup M_2$, such that $\ell(M_1,M_2)>0$. Otherwise $M$ is $\ell_0$-connected. A set $A$ is said to be $\ell_0\mathring{B}$-connected if its intersection with each ball $\mathring{B}(x,R)$ is $\ell_0$-connected. A set $A$ is said to be $\mathring{B}$-closed if its intersection with each ball $\mathring{B}(x,R)$ is a relatively closed subset of $\mathring{B}(x,R)$. Note that a closed set is a fortiori $\mathring{B}$-closed if the unit ball $B(0,1)$ of $X$ is closed. The following result is Theorem 1 of [32]. Theorem A. Any nonempty $\mathring{B}$-closed $\ell_0\mathring{B}$-connected subset of an asymmetric left-complete space $(X,\|\,{\cdot}\,|)$ is $\mathring{B}$-path connected. The next result is a direct consequence of Theorems 7 and A. Corollary 6. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be a $\mathring{B}$-closed, at most countable union of pairwise disjoint closed sets $M_j $ in a left-complete asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then $M$ is not $\ell_0\mathring{B}$-connected. Definition 7. A set $M$ is said to be ${B}$-connected if its intersection with any ball ${B}(x,R)$ is connected. According to Remark 8 in [32], if $M$ is $B$-connected, then $M$ is $\mathring{B}$-connected; moreover, $M$ is $\ell_0\mathring{B}$-connected by Remark 3 in [32]. The following result is Theorem 6 in [32]. Theorem B. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric space whose unit ball is closed, and let $M\subset X$ be a $P$-connected approximatively compact set. Then $M$ is ${B}$-connected. The next result is a direct corollary to Theorem B and Corollary 6. Corollary 7. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an approximatively compact, at most countable union of pairwise disjoint closed sets $M_j$ in a left-complete asymmetric normed linear space $(X,\|\,{\cdot}\,|)$ whose unit ball $B(0,1)$ is closed. Then $M$ is not $P$-connected. The following result is Theorem 2 in [32]. Theorem C. Let $(X,\|\,{\cdot}\,|)$ be a left-complete asymmetric normed linear space and $A \subset X$ be a nonempty $\mathring{B}$-closed set (or let both the set $A$ and the ball $B(0,1)$ be closed). Then the following conditions are equivalent: (a) $A$ is $\mathring{B}$-connected; (b) $A$ is $\mathring{B}$-path-connected; (c) $A$ is $\ell_0\mathring{B}$-connected. Remark 6. Any proximinal set is $\mathring{B}$-closed, and therefore in a left-complete asymmetric normed linear space its $\mathring{B}$-connectedness is equivalent to its $\mathring{B}$-path-connectedness and $\ell_0\mathring{B}$-connectedness. Definition 8. The metric projection $P_M$ is said to be $\ell$-continuous at $x$ if for any sequence $\{x_n\}$ such that $\|x_n-x\|_{\mathrm{sym}}\to 0$. Definition 9. A set $M\subset X$ is said to be $\ell_0P$-connected if $P_M(x)$ is nonempty and $\ell_0$-connected for each $x\in X$. A set $M\subset X$ is said to be $P$-connected if $P_M(x)$ is a nonempty connected set for each $x\in X$. The following result is Theorem 5 in [32]. Theorem D. Let $X$ be a left-complete asymmetric normed linear space and ${M\!\subset\! X}$ be an $\ell_0P$-connected set with $\ell$-continuous metric projection. Then $M$ is $\mathring{B}$-connected. The following result is a consequence of Theorems 7, D, C and Remark 6. Corollary 8. Let $M:=\bigcup_{j\in J}M_j$, $\operatorname{card}J\geqslant 2$, be an $\ell_0P$-connected at most countable union of pairwise disjoint closed subsets $M_j$ of an asymmetric normed linear space $(X,\|\,{\cdot}\,|)$. Then the metric projection onto $M$ is not $\ell$-continuous. 
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