The proof of Theorem 3.4 practically coincides with the proof of
Theorem 3.2, only a few obvious minor changes are needed.
Proof
of Theorem 3.5. The space
X
cannot be continuously mapped
onto the Tychonoff cube
I
ω
1
, since the space
I
ω
1
is not hereditarily normal.
Hence, by a theorem of Shapirovskij in [22], the
π
-character of
X
at some
point
e
2
X
is countable. Since
X
is a strong rotoid, the space
X
is
twistable at
e
. Now it follows from Lemma 3.8 that
e
is a
G
δ
-point in
X
.
Therefore,
X
has a
G
δ
-diagonal in
X
×
X
by Corollary 3.11. Hence,
X
is
metrizable, since
X
is compact.
To prove Theorem 3.7, we need the following elementary fact from [4]:
Propositional 3.12.
If
Z
is a retract of
X
, and
e
2
Z
, and there is a
twister at
e
on
X
, then there is a twister on
Z
at
e
.
Theorem 3.7 immediately follows from the next statement:
Theorem 3.13.
If
Z
is a retract of a strong rotoid
X
, and
Z
is a space
of point-countable type, then
Z
is first-countable at every point
z
at which
the space
Z
has a countable
π
-base.
Proof.
Since
X
is a strong rotoid, Propositions 3.12 and 3.9 imply that
Z
is twistable at every point.
Take any point
z
at which the space
Z
has a countable
π
-base. Now
Lemma 3.8 implies that
z
is a
G
δ
-point in
Z
. Since
Z
is of point-countable
type, it follows that the space
Z
is first-countable at
z
.
Proof
of Theorem 3.6. By a theorem of Shapirovskij in [21], the
π
-
character of
Z
at every point of
Z
is countable, since it doesn’t exceed
the tightness of
Z
. Now Theorem 3.6 follows from Theorem 3.7 (or from
Theorem 3.13).
3.3. Some open questions.
Some of the above results easily generalize
to certain larger classes of spaces. For example, Theorems 3.1–3.5 remain
true for paracompact
p
-spaces, with practically the same proofs.
However, some interesting questions on compact rotoids remain open.
We refer the reader to [6], where many questions on rotoids and close to
them spaces have been formulated. The majority of these questions remain
open.
Problem 3.14.
[6]
Is every compact rotoid a dyadic compactum?
Problem 3.15.
[6]
Is every compact rotoid a Dugundji compactum?
Example 3.16.
It was shown in [10] (Proposition 4.1) that the closed
unit interval
I
= [0
,
1]
is a rotoid. However, the space
[0
,
1]
is not
rectifiable, since it is not homogeneous. Therefore, not every compact
rotoid is rectifiable. We will now strengthen this conclusion. Put
X
=
I
ω
.
Clearly,
X
is also a rotoid. But
X
is also homogeneous. Therefore,
X
is a
strong rotoid. However,
X
is not rectifiable, as it was shown in [15].
The next question is closely related to Problems 3.14 and 3.15.
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2013. № 2
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