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\begin{document}


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\noindent{
\label{beginninofthedocument}
\begin{picture}(150,36)
\put(0,8){\includegraphics{../tplogo09.eps}}
\put(1,19){\tiny{Volume 59 (2022)}}
\put(1,12){\tiny{Pages
\pageref{beginninofthedocument}-\pageref{endofthedocument}}}
%\put(1,64){\tiny{\textbf{http://topology.auburn.edu/tp/}}}
\put(1,58){\tiny{\textbf{http://topology.nipissingu.ca/tp/}}}
\put(1,0){\tiny{E-Published on  April 25, 2021}}
\end{picture}
\vspace{0.5in}



\renewcommand{\bf}{\bfseries}
\renewcommand{\sc}{\scshape}

%\null
\vspace{0.5in}
\title[On metrizable subsets of hereditarily normal spaces...]{On metrizable subsets of hereditarily normal compact spaces}

%    Information for first author:
\author{Heikki Junnila}
\address[Junnila]{Department of Mathematics and Statistics,\hfill\break 
\hspace*{1.738cm}University of Helsinki, 00140 Helsinki, Finland}
\email{heikki.junnila@helsinki.fi}

\author{Kazuo Tomoyasu}
\address[Tomoyasu]{General Education, National Institute of Technology,\hfill\break \hspace*{2.08cm}Miyakonojo College, Miyakonojo-shi Miyazaki 885-8567,\hfill\break \hspace*{1.96cm} Japan}
\email{tomoyasu@cc.miyakonojo-nct.ac.jp\label{endofthedocument}}

\subjclass{54D35, 54D15, 54D05}
\keywords{hereditarily normal, $\omega_1$-compact, rim-Lindel\"of}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}

\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}




\begin{abstract}
Let $X$ be a metrizable space which has a
hereditarily normal $\omega_1$-compactification.
We show that $X$ is rim-separable and that if $X$
is also connected, then $w(X)\leq\omega_1$ and $X$
has a $\sigma$-point-finite base by sets with
separable boundaries.
\end{abstract}

\thanks{\copyright 2021 Topology Proceedings}
\maketitle

\medskip


\section{\bf Introduction} One of the finest contributions of Philip Zenor
to the theory of point-set topology was the introduction of the concept of
{\it monotone normality} in 1970 ([13]). This concept turned out to be
extremely fruitful and some of the most spectacular results in topology in
the last decades deal with monotonically normal spaces.

Also this paper has its roots in work on monotone normality. It is known
that many metrizable spaces have monotonically normal compactifications.
For example, every locally separable metrizable space has a hereditarily
paracompact monotonically normal compactification (see the proof of
Proposition 3 below) and this kind of a
compactification exists also for every strongly zero-dimensional
metrizable space (see [8]). A problem
due to P. Gartside  asked whether every metrizable
space has a monotonically normal compactification ([3, Problem 10]). The problem was solved
in [9] by the result that the hedgehog-space
$J(\omega_1)$ admits no monotonically normal compactification.

The result of [9] mentioned above was actually stronger: $J(\omega_1)$ has
no hereditarily normal $\omega_1$-compactification. Recall that
a space $X$ is $\omega_1$-{\it compact} provided that every
closed discrete subset of $X$ is countable. Every Lindel\" of
space and every countably compact space is $\omega_1$-compact.

In this paper, we study properties of those metrizable spaces $M$ which can be
embedded into hereditarily normal $\omega_1$-compact spaces. We show that
such a space $M$ is rim-separable.
In particular, we consider the
following problem raised in [9].
\medskip

\noindent{\bf Problem 1.} {\rm [9]} {\it Let $M$ be a connected metrizable space with a
hereditarily normal compactification. Is $M$ separable?}
\smallskip


The problem is still open, but we show that, for
$M$ as above, the weight of $M$ is at most $\omega_1$
and $M$ has a $\sigma$-point-finite base by sets with
separable boundaries. Moreover, we show that if $M$ is
also locally connected, then $M$ is separable.
\medskip

For the meaning of terms used without definition in this paper, see [4].
\medskip


\section{\bf Some properties of hereditarily normal spaces}
The results on spaces with hereditarily normal
$\omega_1$-compactifications in [9] relied on the following
property of hereditarily normal spaces. Here we note
that this property actually characterizes hereditary normality.
\medskip


\noindent{\bf Lemma 1.} {\it A topological space $Z$ is hereditarily normal if, and
only if, for all subsets $A$ and $S$ of $Z$ such that $S$ is closed
and $A\cap S=\emptyset$, there exists an open $V\subset Z$
with $A\sis V$ and $\overline{V}\cap S\sis\overline{A}$}.}
\smallskip

\begin{proof} {\it Necessity} is proved in [9, Lemma 3].
\smallskip

\noindent{\it Sufficiency}.
We will first show that $Z$ is normal. Let $A$ and $B$ be
disjoint closed subsets of $Z$. By the assumption, there
exists an open set $V\subset Z$ such that $A\subset V$ and
$\overline{V}\cap B\subset\overline{A}=A$. Hence
$\overline{V}\cap B\subset A\cap B=\emptyset$, and thus $B\subset
Z\sm\overline{V}$. Now $V$ and $Z\sm\overline{V}$ are
disjoint open sets containing $A$ and $B$ respectively. We have
shown that $Z$ is normal. Next, we note that the property considered is
hereditary with respect to open subspaces. It follows that every
open subspace is normal, and hence that $Z$ is hereditarily normal
(see [4, Theorem 2.1.7]).
\end{proof}
\smallskip

Recall that a family $\mathcal{L}$ of subsets of a topological space $X$
is {\it discrete} provided that every point of
$X$ has a neighbourhood which meets at most one member
of the family $\mathcal{L}$. The family $\mathcal{L}$ is
{\it relatively discrete} provided that $\mathcal{L}$ is a discrete family
in the subspace $\bigcup\mathcal{L}$ of $X$.

With the help of Lemma 1, we can derive the following variant of
[9, Lemma 4]. We state the result in terms of cardinal
functions. Recall that the {\it spread} s(X) (the {\it extent}
$e(X)$) of a topological space $X$ is the supremum of the
cardinalities of discrete (closed discrete) subsets of $X$.
\medskip

\noindent{\bf Lemma 2.} {\it Let $U$ be an open subspace of a hereditarily normal space
$Z$ and let $\mathcal{L}$ be a relatively discrete family in $Z$ such that, for
every $L\in\mathcal{L}$, the set $L\cap U$ is non-closed in $L$. Then the
inequality $|\mathcal{L}|\leq e(U)$ holds.}
\smallskip

\begin{proof} Denote by $G$ the union of all those open subsets of $Z$ which
meet at most one member of $\mathcal{L}$. Note that $G$ is open,
$\bigcup\mathcal{L}\sis G$ and $\mathcal{L}$ is a discrete family in $G$. Let
$A=G\sm U$.
By Lemma 1, there exists an open set $V$ such that $A\subset V$ and
$\overline{V}\pois G\sis\overline{A}$. For every $L\in\mathcal{L}$,
since $L\sm U\sis L\cap V$ and since the set
$L\cap U$ is not closed in $L$, there exists a point
$x_L\in L\cap V\cap U$.

Denote by $E$ the subset $\{x_L: L\in\mathcal{L}\}$ of $V\cap U$,
and note that we have
$$\overline{E}\sm G\subset\overline{V}\sm
G\subset\overline{A}\subset Z\sm U\,.$$
It follows that $\overline{E}\cap U\subset G$. Since
$\mathcal{L}$ is a discrete family in $G$, the set $G\cap E$ is closed and
discrete in $G$. It follows from the foregoing that we have
$\overline{E}\cap U\subset\overline{E}\cap G=E\cap G\subset E$.
Hence $E$ is closed in $U$ and it follows, since $E$ is relatively
discrete, that $|e(U)|\geq |E|=|\mathcal{L}|$.\end{proof}
\smallskip

In the sequel, we shall use the following simple consequence of
Lemma 2.
\medskip

\noindent{\bf Corollary 1.} {\it Let $U$ be an open subspace of a hereditarily normal space
$Z$ and let $\mathcal{G}$ be a disjoint family of open subsets of $Z$ such
that we have $G\cap\partial U\ne\emptyset$ for every $G\in\mathcal{G}$.
Then the inequality $|\mathcal{G}|\leq e(U)$ holds.}
\medskip



\begin{proof} A disjoint open family is relatively discrete, and for an
open set $G$, the condition $G\cap\partial U\ne\emptyset$ is
equivalent with the condition that the set $G\cap U$ fails to be
closed in $G$. \end{proof}



Corollary 1 is useful in connection with the
property of ``hereditary collectionwise Hausdorffness'': recall
that a topological space $X$ is {\it collectionwise Hausdorff}
provided that, for every closed discrete subset $A$ of $X$, we
can assign neighbourhoods $V_a$ for all points $a\in A$ so that
$V_a\cap V_b=\emptyset$ whenever $a,b\in A$ are distinct. It is
easy to see that $X$ satisfies this property hereditarily if, and
only if, the stated conclusion obtains for every {\it relatively} discrete subset $A$ of $X$.
\medskip

\noindent{\bf Lemma 3.} {\it Let $Z$ be a hereditarily normal space and $X$ a
dense hereditary collectionwise Hausdorff subspace of $Z$. Then the
inequality $s(X\cap\partial U)\leq e(U)$ obtains for every open
$U\subset Z$.}
\medskip

\begin{proof} Let $U$ be an open subspace of $Z$. To prove the inequality
stated in the lemma, let $A$ be a relatively discrete subset of
$X\cap\partial U$.
Since $X$ is hereditarily collectionwise Hausdorff, we can find
disjoint open neighbourhoods $N_a$ for the points of $A$ in the
subspace $X$. For every $a\in A$, let $V_a$ be an open subset of
$Z$ such that $V_a\cap X=N_a$. Since $X$ is dense in $Z$, the
sets $V_a$, $a\in A$, are disjoint.
Corollary 1 now shows that we have $|A|=|\{V_a: a\in A\}|\leq e(U)$.
\end{proof}

A similar result is implicitly contained in [10, proof
of Lemma 1.2].

If we make more assumptions on the space and the open set, we can
strengthen the above conclusion.
\medskip

\noindent{\bf Corollary 2.} {\it Let $Z$ be a hereditarily normal $\omega_1$-compact space
and $X$ a dense hereditary collectionwise Hausdorff subspace of $Z$.
Then the subspace $X\cap\partial U$ has countable spread whenever
$U$ is an open $F_\sigma$-subset of $Z$.}


\begin{proof} $Z$ is $\omega_1$-compact means that $e(Z)\leq\omega$,
and this inequality is inherited by all closed subsets and hence
by all $F_\sigma$-subsets of $Z$.  \end{proof}

\renewcommand{\bf}{\bfseries}

\section{\bf Spaces with $\sigma$-additively rim-Lindel\"of bases}

We can use Corollary 2 to show that every metrizable space with a
hereditarily normal $\omega_1$-compactification is {\it
rim-separable}, i.e., it has a base by sets with separable
boundaries.

We call a base $\mathcal{B}$ of a space $X$ a {\it rim-P base} if $\partial B$ has property $P$
for each $B\in \mathcal{B}$. If $\partial\bigcup\mathcal{C}$ has property $P$ for each countable
$\mathcal{C}\sis\mathcal{B}$, then we say that $\mathcal{B}$ is a {\it $\sigma$-additively rim-P base}.

If $X$ has a rim-P base, we say that $X$ is a {\it rim-{\rm P} space}.

We abbreviate {\it rim-hereditarily-Lindel\"of} to
{\it rim-hL}.

For (a base of) a metrizable space, the properties ``rim-Lindel\"of'',
``rim-separable'' and ``rim-hL'' are mutually equivalent.


In the rest of this paper, we will study spaces with a $\sigma$-additively
rim-Lindel\"of base. We start by observing that these spaces share a
property of paracompact spaces.
\medskip


\noindent{\bf Proposition 1}. {\it Assume that $X$ has a $\sigma$-additively rim-Lindel\"of base.
Then the closure of each Lindel\"of subset of $X$ is Lindel\"of.}


\begin{proof} Let $\mathcal{B}$ be a $\sigma$-additively rim-Lindel\"of base of $X$, and let $L\sis X$
be Lindel\"of. To show that $\ol{L}$ is Lindel\"of, let $\mathcal{G}$ be an open cover of $X$.
Set $\mathcal{C}=\{B\in\mathcal{B}: B\sis G\txt{ for some }G\in\mathcal{G}\}$, and note that
$\mathcal{C}$ covers $X$. As $L$ is Lindel\"of, there exists a countable
$\mathcal{C}'\sis\mathcal{C}$ such that $L\sis\bigcup\mathcal{C}'$. Since $\mathcal{C}'$ is a countable
subfamily of $\mathcal{B}$, the set $\partial\bigcup\mathcal{C}'$ is Lindel\"of. As a
consequence, there exists a countable $\mathcal{C}''\sis\mathcal{C}$ such that
$\partial\bigcup\mathcal{C}'\sis\bigcup\mathcal{C}''$. Now $\mathcal{C}'\cup\mathcal{C}''$ is a countable
subfamily of $\mathcal{C}$ and
$$\ol{L}\sis\ol{\bigcup\mathcal{C}'}=\bigcup\mathcal{C}'\cup\partial\bigcup\mathcal{C}'
\sis\bigcup\big(\mathcal{C}'\cup\mathcal{C}''\big)\,.$$
It follows that some countable subfamily of $\mathcal{G}$ covers $\ol{L}$.
\end{proof}

We will soon provide examples of metrizable spaces which do not have a
$\sigma$-additively rim-Lindel\"of base. Hence the condition, that each
Lindel\"of subset has Lindel\"of closure, is not sufficient for the
existence of a $\sigma$-additively rim-Lindel\"of base.
However, it is easy to see that a locally Lindel\"of space has a $\sigma$-
additively rim-Lindel\"of base provided that each Lindel\"of subset of
the space has Lindel\"of closure.

 Proposition 1 can be used to find spaces without $\sigma$-additively
rim-Lindel\"of bases. For example, we see that every non-Lindel\"of separable
space fails to have such a base.

The next result can be used to indicate spaces with a
$\sigma$-additively rim-Lindel\"of base.
\medskip

\noindent{\bf Theorem 1}. {\it If a hereditarily paracompact space has a hereditarily normal
$\omega_1$- compactification, then the space has a $\sigma$-additively
rim-hL base}.

\begin{proof} Let $Z$ be a hereditarily normal $\omega_1$-compactification of the
hereditarily paracompact space $X$. Every paracompact space is collectionwise
Hausdorff and every hereditarily paracompact space of countable spread is
hereditarily Lindel\" of.
It follows, by Corollary 2, that the family
$\mathcal{B}=\{G\cap X: G\txt{ is an open }F_\sigma\txt{ subset of }Z\}$
consists of rim-hL subsets of $X$.
Since $Z$ is a Tychonoff space, $\mathcal{B}$ is a base of $X$. Moreover, $\mathcal{B}$ is
closed under countable unions, so $\mathcal{B}$ is a $\sigma$-additively rim-hL
base.  \end{proof}





\noindent{\bf Corollary 3}. {\it Every hereditarily paracompact Lindel\"of space is
rim-hL}.
\medskip



Theorem 1 shows that every metrizable space with a hereditarily normal
$\omega_1$-compacti\-fi\-cation has a $\sigma$-additively rim-hL base and therefore a
 $\sigma$-additively rim-separable base. In [8, Example 5.21] it is shown that
every strongly zero-dimensional metrizable space has a hereditarily
paracompact compactification. Hence every strongly zero-dimensional
metrizable space has a $\sigma$-additively rim-separable base. This result,
however, has an easy direct proof. Let $X$ be strongly
zero-dimensional metrizable space. It is well known that $X$ is a
{\it non-Archimedean space}, i.e., $X$ has a {\it point-monotone} base $\mathcal{B}$
(that is, for every $x\in X$, the family $\{B\in\mathcal{B}: x\in B\}$ is monotone).
We can assume that $\es\not\in\mathcal{B}$. For every $B\in\mathcal{B}$, choose a point $x_B\in B$.
Then it is easy to see that for every $\mathcal{C}\sis\mathcal{B}$, we have
$\partial\bigcup\mathcal{C}\sis\overline{\{x_B: B\in\mathcal{C}\}}$, and it follows that
$\partial\bigcup\mathcal{C}$ is separable if $\mathcal{C}$ is countable.

The following particular consequence of Theorem 1
shows that there are many metrizable spaces
which do {\it not} have hereditarily normal compactifications.
\medskip

\noindent{\bf Corollary 4}. {\it A metrizable space with a hereditarily normal $\omega_1$-
compactification is rim-separable.}
\smallskip

Examples of metrizable spaces which fail to be rim-separable are provided,
e.g., by hedgehog-spaces $J(\kappa)$, where $\kappa$ is uncountable, and by
non-separable normed spaces.
\medskip

F.B. Jones has shown in [6] that every connected and locally connected
rim-separable metrizable space is separable. By this result and Corollary 4,
we obtain the following partial solution to Problem 1.
\medskip

\noindent{\bf Proposition 2.} {\it Let $X$ be a connected and locally connected metrizable space with a
hereditarily normal $\omega_1$-compactification. Then $X$ is separable}.
\vspace*{-2mm}

Our next result gives a partial solution to [9, Problem 2].
\medskip

\noindent{\bf Proposition 3.} {\it The following are equivalent for a locally connected metrizable space $X$:

\noin{\rm\bf A.} $X$ has a hereditarily normal $\omega_1$-compactification.

\noin{\rm\bf B.} $X$ has a hereditarily paracompact monotonically normal compactification.

\noin{\rm\bf C.} $X$ is locally separable.}


\begin{proof} ${\mathbf A}\Rightarrow {\mathbf C}{\mathbf :}$ Assume that {\bf A} holds. By Proposition
2, every connected component of $X$ is separable. Hence {\bf C} holds.
\smallskip

The implication ${\mathbf B}\Rightarrow {\mathbf A}$ holds trivially. To prove that
${\mathbf C}\Rightarrow {\mathbf B}$, assume that
$X$ is locally separable. By a result of P.S. Alexandroff [1], $X$ is the direct
sum of a family of separable spaces. Each summand
has a metrizable compactification, and hence $X$ embeds in a direct sum $Y$
of compact metrizable spaces. By [2, Theorem 3.5], the one-point compactification
$\ol{Y}$ of $Y$ is
monotonically normal. It is easy to see that $\ol{Y}$ is also
hereditarily paracompact.  \end{proof}


In 1954, Jones [7] raised the problem whether every connected
rim-separable metrizable space is separable. Corollary 4 above shows that
a positive solution to Jones' problem would give a positive solution to
Problem 1. However, the problem was solved in the negative in 1960 by
L.B. Treybig [12], with an example of a non-separable connected
rim-separable metrizable space $\Sigma$. In 1970, P. Roy [11] constructed such a
space $\Gamma$ which is even rim-compact. In $\Sigma$ (in $\Gamma$), the set of
points without a separable (compact) neighborhood is separable. By a result of
T. Hoshina ([5]), $\Gamma$ has a countable points compactification, i.e., a
Hausdorff compactification $K$ with $|K\pois\Gamma|\leq\omega$.

The non-separable, rim-separable connected spaces $\Sigma$ and $\Gamma$ do
not solve Problem 1. To see this, we show that neither space
has a $\sigma$-additively rim-separable base. Let $S$ be either $\Sigma$ or
$\Gamma$ and assume on the contrary that $S$ has a
$\sigma$-additively rim-separable base $\mathcal{B}$. Let $T$ be the (separable) subset of $S$
consisting of all points without a separable neighborhood.
Let $U$ be a neighborhood of the set $T$ in $S$. Some countable
subfamily $\mathcal{C}$ of the family $\mathcal{B}'=\{B\in\mathcal{B}: B\sis U\}$ covers the
separable set $T$. Let $O=\bigcup\mathcal{C}$ and note that $O$ is open, $T\sis
O\sis U$ and $\partial O$ is separable. This shows that, in the terminology
of [11], the space $S$ is {\it locally peripherally separable about $T$}.
The space $S$ and the subset $T$ satisfy all assumptions of [11, Theorem 3.1]
and it follows that $S$ is separable. This contradiction shows that $S$ does
not have a $\sigma$-additively rim-separable base. By Theorem 1, $S$ does
not have a hereditarily normal $\omega_1$-compactification.
\smallskip

Recall that $X$ has {\it countable tightness} provided
that, for every $A\sis X$, we have
$\overline{A}=\bigcup\{\overline{B}: B\sis A\txt{ and }B\txt{ is countable}\}$.
Every first countable space, in particular, every metrizable
space, has countable tightness.
\medskip

\noindent{\bf Proposition 4}. {\it Let $X$ be a connected space of countable tightness with a
 $\sigma$-additively rim-Lindel\"of base.
Then every open cover of $X$ has a subcover of size $\leq\omega_1$}.

\begin{proof} Let $\mathcal{B}$ be a $\sigma$-additively rim-Lindel\"of base of $X$. It suffices to show that
every cover $\mathcal{C}\sis\mathcal{B}$ has a subcover of size $\leq\omega_1$. Let $\mathcal{C}\sis\mathcal{B}$
cover $X$. We use transfinite induction to define
a sequence $\langle\mathcal{C}_\alpha\rangle_{\alpha<\omega_1}$
of countable subfamilies of $\mathcal{C}$ as follows.

Let $\mathcal{C}_0=\{B\}$, where $B\in\mathcal{C}$ is non-empty.
If $\alpha<\omega_1$ is such that $\mathcal{C}_\beta$ has been
defined for each $\beta<\alpha$,
then we set $D_\alpha=\bigcup_{\beta<\alpha}\bigcup\mathcal{C}_\beta$,
and we note that $\partial D_\alpha$ is Lindel\" of. As a consequence,
there exists a countable $\mathcal{C}_\alpha\subset\mathcal{C}$
such that $\partial D_\alpha\subset\bigcup\mathcal{C}_\alpha$.
This completes the induction.


Note that the sequence $\langle D_\alpha\rangle_{\alpha<\omega_1}$
of open sets is {\it strongly monotone} in the sense that
$\overline{D_\beta}\subset D_\alpha$ whenever $\beta<\alpha$.
It follows, by countable tightness of $X$, that
$D=\bigcup_{\alpha<\omega_1}D_\alpha$ is clopen.
Since $D\ne\emptyset$ and $X$ is connected, we have that $D=X$.
As a consequence, the family
$\mathcal{C}^{\prime}=\bigcup_{\alpha<\omega_1}\mathcal{C}_\alpha$
covers $X$.
Moreover, we have that $\mathcal{C}^{\prime}\subset\mathcal{C}$ and
$|\mathcal{C}^{\prime}|\leq\omega_1$.\end{proof}


Recall that the {\it weight} $w(X)$ of a space $X$
is the least cardinality of a base of $X$.
Since a metrizable space $X$ is separable iff $w(X)\leq\omega$,
the following result, together with Theorem 1, gives a partial solution
to Problem 1.
\medskip

\noindent{\bf Corollary 5}. {\it Let $X$ be a connected metrizable space with a
$\sigma$-additively rim-separable base. Then $w(X)\leq\omega_1$}.
\smallskip

In our final theorem, we show that a connected metrizable space
with a $\sigma$-additively rim-separable base has a $\sigma$-point-finite
rim-separable base. We need an auxiliary result.
\medskip

\noindent{\bf Lemma 4.} {\it Let $X$ be a  metrizable space
with a $\sigma$-additively rim-separable base $\mathcal{B}$,
and let $\mathcal{C}\sis\mathcal{B}$ be such that $|\mathcal{C}|\leq\omega_1$.
Then $\mathcal{C}$ has a point-finite refinement by rim-separable open
sets}.





\begin{proof} Write $\mathcal{C}=\{B_\alpha: \alpha<\lambda\}$, where
$\lambda\leq\omega_1$.
Let $d$ be a compatible metric for $X$ with $d\leq 1$.
For every $\alpha<\lambda$, the set
$$U_{\alpha}=\{x\in X: d(x, X\sm B_\alpha)>2d(x,
X\sm\bigcup_{\beta<\alpha}B_\beta)\}$$
\vspace*{-4mm}

 \noin is open and
$B_\alpha\sm\bigcup_{\beta<\alpha}B_\beta \subset
U_\alpha\subset B_\alpha$.

The family $\{U_\alpha: \alpha<\lambda\}$
is an open refinement of $\mathcal{C}$.
Moreover, this family is point-finite.
Otherwise there would exist $x\in X$ and
$\alpha_1<\alpha_2<\cdots <\lambda$ such that
$x\in U_{\alpha_n}$ for every $n$.
Then we would have $d(x,X\sm B_{\alpha_{n+1}})>
2d(x,X\sm\bigcup_{\beta<\alpha_{n+1}}B_\beta)\geq 2d(x,X\sm B_{\alpha_n})$
for every $n$. As a consequence, we would have
$d(x,X\sm B_{\alpha_{n+1}})>2^{n}d(x,X\sm B_{\alpha_{1}})$
for each $n$. Since $d(x,X\sm B_{\alpha_1})>0$, this would contradict
the assumption that $d\leq 1$.

To complete the proof, let $\alpha<\lambda$, and note that
$\partial\bigcup_{\beta<\alpha}B_\beta$ is separable. Since
$B_\alpha\cap\partial\bigcup_{\beta<\alpha}B_\beta$
$\subset B_\alpha\sm\bigcup_{\beta<\alpha}B_\beta$
$ \subset U_\alpha$,
there exists a countable $\mathcal{N}\sis\mathcal{B}$ such
that $B_\alpha\cap\partial\bigcup_{\beta<\alpha}B_\beta$
$\subset\bigcup\mathcal{N}\subset U_\alpha$.
Since
$B_\alpha\cap\partial\bigcup_{\beta<\alpha}B_\beta=
B_\alpha\cap\partial\big(B_\alpha\sm
\bigcup_{\beta<\alpha}B_\beta\big)$,
we see that the set
$V_\alpha=\big(B_\alpha\sm
\bigcup_{\beta<\alpha}B_\beta\big)\cup\bigcup\mathcal{N}$
is open.
Moreover, we have that
$B_\alpha\sm\bigcup_{\beta<\alpha}B_\beta\subset
V_\alpha\subset U_\alpha$
and $\partial V_\alpha$ is
contained in the union of the separable sets
$\partial B_\alpha$ and $\partial\bigcup\mathcal{N}$.

 By the foregoing, $\{V_\alpha: \alpha<\lambda\}$ is a
point-finite refinement of $\mathcal{E}$
by open sets with separable
boundaries. \end{proof}


It can be shown that the point-finite open refinement constructed above for the
family $\mathcal{C}$ is locally countable in the subspace $\bigcup\mathcal{C}$, but the following
problem is open.
\medskip

\noindent{\bf Problem 2.} {\it Let $X$ be a  metrizable space with a $\sigma$-additively rim-separable base
$\mathcal{B}$, and let $\mathcal{C}\sis\mathcal{B}$ be a cover of $X$ such that $|\mathcal{C}|\leq\omega_1$.
Does $\mathcal{C}$ have a locally finite refinement by rim-separable open sets?}
\medskip

A positive solution to Problem 2 would give a positive solution to Problem 1.
This is a consequence of the following observation. Let $\mathcal{U}=\{U_\alpha:
\alpha<\omega_1\}$ be a locally finite cover of a space $X$ by rim-separable
open sets. For every $\alpha<\omega_1$, let
$V_\alpha=U_\alpha\sm\bigcup\{U_\beta: \beta<\alpha\txt{ and
}U_\alpha\cap\partial U_\beta=\es\}$. Then the family $\mathcal{V}=\{V_\alpha:
\alpha<\omega_1\}$ is a star-countable open refinement of $\mathcal{U}$.
Now if $X$ is connected, metrizable and has a hereditarily normal
$\omega_1$-compactification, then our earlier results show that
$X$ has a $\sigma$-additively rim-separable base and $w(X)\leq\omega_1$; the
preceding argument would give a star-countable open refinement for
every open cover of $X$ and it is well known (see, e.g., [4, Lemma 5.3.9])
that a star-countable open cover of a connected space is countable.
\medskip


\noindent{\bf Theorem 2}. {\it Let $X$ be a connected metrizable space
with a $\sigma$-additively rim-separable base.
Then $X$ has a $\sigma$-point-finite rim-separable base.}



\begin{proof} Let $\mathcal{B}$ be a $\sigma$-additively rim-separable base of $X$.
By Corollary 5, we have $w(X)\leq\omega_1$. Let $d$ be a
compatible metric for $X$. For every $n\in\N$, there exists
a cover $\mathcal{U}_n$ such that $\mathcal{U}_n\subset\mathcal{B}$, $|\mathcal{U}_n|\leq\omega_1$
and $d$-diam$U\leq{1\over n}$ for every $U\in\mathcal{U}_n$.
By Lemma 4, each $\mathcal{U}_n$ has a point-finite refinement
$\mathcal{V}_n$ by rim-separable open sets. The family
$\bigcup_{n\in\N}\mathcal{V}_n$ is a $\sigma$-point-finite rim-separable base
of $X$. \end{proof}

\noindent{\bf Problem 3.} {\it Is it possible to construct the $\sigma$-point-finite
base to be $\sigma$-additively rim-separable?}
\medskip


Theorem 2 does not solve Problem 1, because there exists a non-separable connected
metrizable space with a $\sigma$-point-finite rim-separable base. To see this,
consider Treybig's space $\Sigma$. This space contains a closed separable subset
$T$ such that the subspace $\Sigma\sm T$ is locally separable. Since $\Sigma$
is rim-separable, there is a countable family $\mathcal{B}_0$ of rim-separable open sets
such that $\mathcal{B}_0$ contains a base for every $t\in T$. The locally separable
open subspace $\Sigma\sm T$ has a base $\mathcal{B}_1$ by separable sets such that
$\mathcal{B}_1$ is $\sigma$-discrete in $\Sigma$. The family
$\mathcal{B}_0\cup\mathcal{B}_1$ is a $\sigma$-discrete base for $\Sigma$ by rim-separable
sets.

Since Roy's space $\Gamma$ is rim-compact and it has a locally compact open
subspace whose complement is separable, the above argument on $\Sigma$ can
be applied on $\Gamma$ to show that the latter space has a $\sigma$-discrete
rim-compact base.






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\end{document}


