数理解析研究所講究録
第 1987 巻 2016 年 32-36

32

NOTES ON ELEMENTARY SUBMODEL SPACES
TOSmMICHI USUBA

1. INTRODUCTION
Throughout this note, will denote a suﬃciently large regular cardinal.
will$X=\langle X,$
denote the set of all sets with hereditariry cardinality
will
denote a topological space, and $M$ an elementary submodel ofwith $X\in M.$
The elementary submodel space was introduced and studied in Junqueir$*$ Tall
[1].
$\theta$$\mathcal{H}_{\theta}$$<\theta.$$x$$\tau\rangle$$\mathcal{H}_{\theta}$

Definition 1.1. For $X$ and $M$ , the elementary submodel space$X\cap M$ with topology generated by the family $\{O\cap M : O\in\tau\}.$$X_{M}$

is the space

The elementary submodel spaces reflect many properties of the original spaces
and vise verse, for instance:

Fact 1.2 ([1]).
(1) $(n=0,1,2,3,3 \frac{1}{2})X$ is(2) Ifis compact then $X$ is compact.

$T_{n}\Leftarrow\Rightarrow X_{M}$is$T_{n}.$$X_{M}$

In [2], Tall showed that ifsatisfies some -like properties, then$X$
the same to the original space :$X_{M}$$\mathbb{R}$$X_{M}$

must beTheorem 1.3 (Tall [2]). If
is locally compact, metrizable, separable, and uncountable, then $X_{M}=X$ . In particular if
is homeomorphic to the realthen $X_{M}=X.$$X_{M}$$X_{M}$$hne\mathbb{R}_{i}$Among this result, Tall asked the following question:
Question 1.4 (An irrational problem). Suppose $X_{M}$ is completely metrizable, separable, and uncountable. Does $X_{M}=X$ ? Or, if $X_{M}$ is homeomorphic to the, does $X_{M}=X^{q}$imationals$\mathbb{R}\backslash \mathbb{Q}$In [2] and [3], Tall gave some partial answers of an irrational problem, and it
turned out that if
does not exist, then the aﬃrmative answer of an irrational
problem holds, so it is consistent with ZFC. However it is unknown whether thenegative answer of an irrational question is consistent with ZFC. In this notes, wewill give another partial answer of an irrational problem under certain cardinalarithmetic assumption.
$o\#$33. For
every cardinalandTheorem 1.5. Suppose
is completely metrizable, separable, and uncountable, thenevery space $X$ , if$2^{\kappa}=\kappa^{+}for$$2^{\omega}>\omega_{1}$$\kappa\geq 2^{\omega}$$X_{M}$$X=X_{M}.$2. PROOFS
in $X$ is leftLet $X$ be a topological space. Recall that a sequenceof open:separated (right-separated, respectively) if there is a sequence
for every $j<i$ $(j>i,$ respectively) . The
butsets such that
following is well-known:$\langle x_{i}:i<’\delta\rangle$$\langle O_{i}$$x_{i}\in O_{i}$$i<\delta\rangle$$x_{j}\not\in O_{i}$Fact 2.1. Let be a regular Cardinal.
(1) $X$ has a left-separated sequence of length if and only if there is a subset
of $X$ such that $Y$ has no subset of $size<\kappa$ which is dense in $Y.$(2) $X$ has a right-separated sequence of length if and only if there are a subset
covers but every subfamily$YofX$ and a family of open sets suchof of $size<\kappa$ does not cover $Y.$$\kappa$$\kappa$$Y$$\kappa$$Y$$that\mathcal{U}$$\mathcal{U}$$\mathcal{U}$The following lemmas may be folklores, but we prove it for the reader’s convenience.
Lemma 2.2. Let$\kappa$

be a regular cardinal. Let $X$ be a topological space and supposethat :(1) For every subset $Y$ and open

coveru of , there is a subcover$Y$

which covers , and(2) For every point $x\in X$ , there is a family
andThen$of\mathcal{U}$of size$Y$$<\kappa$$\mathcal{U}$of open sets such that$|\mathcal{U}|<\kappa$$\{x\}=\cap \mathcal{U}.$$|X|\leq 2^{<\kappa}.$Proof. Take

$M\prec \mathcal{H}_{\theta}$such that$\kappa,$$X\in M,$ $|M|=2^{<\kappa},$ $\kappa\subseteq M$ , and $<\kappa M\subseteq M$. It

is enough to see that
. For $y\in X\cap M$ , take. Fix
Suppose, to the contrary, that. Note thatandwitha family of open sets
. The family $\{O_{y} : yEX\cap M\}$ is anwith
Then we can findsuch
open cover of $X\cap M$ . By our assumption, there is. Then, by, we haveandcovers $X\cap M$ . Since
that$X$for
covers the whole of . This contradicts thatthe elementarity of $M,$
$X\underline{\subseteq}M.$$x^{*}\in X\backslash M$$X\not\subset M$$\mathcal{U}_{y}\in M$$O_{y}\in \mathcal{U}_{y}\cap M$$\mathcal{U}_{y}\underline{C}M.$$|\mathcal{U}_{y}|<\kappa$$\{y\}=\cap \mathcal{U}_{y}$$x^{*}\not\in O_{y}$$\mathcal{U}’\subseteq\{O_{y} :$|\mathcal{U}’|<\kappa$$\mathcal{U}’\in M$$\mathcal{U}’\subseteq M$$\mathcal{U}’$$x^{*}\not\in O$$\mathcal{U}’$every

y\in X\cap M\}$$\square$$O\in \mathcal{U}’.$Lemma 2.3. Let rc be a regular cardinal. Let $X$ be a -space such that for every
subset $Y$ and open cover of $Y$ , there is a subcover of of $size<\kappa u,hich$ covers
Y. Then$T_{2}$$\mathcal{U}$$|X|\leq 2^{<\kappa}.$$\mathcal{U}$34

Proof. By Lemma 2.2, it is enough to see that for every $x\in X$ , there is a family$\mathcal{U}$andof open sets of such that$x\inX$for every with size
Letand suppose to the contrary thatLet
be the family of all open neighborhoods of and ノ. Since every open coverof size
By our assumption,for every, we have that. On the other hand,ofhas a subcover of size$X$$y\neqx$, there is an open neighborhood ofis , for each $y\in X$ withsince, a contradiction.
with
Thus
$\{x\}=\cap \mathcal{U}.$$|\mathcal{U}|<\kappa$$\{x\}\neq\cap \mathcal{U}$$=$$x$$\mathcal{U}_{0}$$<\kappa$$X\backslash \{x\}$$\{\overline{O}\backslash \{x\} :O\in \mathcal{U}_{0}\}.$$<\kappa$$\mathcal{F}’\subseteq \mathcal{F}$$\cap \mathcal{F}’\neq\emptyset$$<\kappa.$$\mathcal{U}$$\cap \mathcal{F}\neq\emptyset$$O$$T_{2}$$y\not\in\overline{O}.$$x$$\square$$\cap \mathcal{F}=\emptyset$

is uncountable, second countable, and . Let $\delta\in M$
Lemma 2.4. Supposebe the ordinal with $ot(M\cap\delta)=\omega_{1}$ (note that $|M\cap|X||\geq\omega_{1}$ , hence such a must
, and for every subset $Y$ of $X$ , theexist). Then is regular uncountable,
following hold:$X_{M}$$T_{2}$$\delta$$\delta\leq|X|$$\delta$(1) $Y$ has a subset which is dense in $Y$ and of $size<\delta$ , and(2) every open cover of has a subcover of $size<\delta.$
$Y$

Hence

$|X|\leq 2^{<\delta}$. Moreover if$X_{M}$

is$T_{3}$

and$2^{<\delta}=\delta$, lhen $X$ has an open base of$size<\delta.$. Then there is a cofinal map :$M,$$f((cf(\delta)\cap M)$ is a cofinal map
with $f\in M$ . By the elementarity of, we have $ot(M\cap cf(\delta))<ot(M\cap\delta)=\omega_{1}.$. Since
fromtoThus $f(M\cap cf(\delta))$ must be bounded in $\sup(M\cap\delta)$ , this is a contradiction.
. By
For (1), suppose $X$ has a subset which has no dense subset of size
$X$in . Let
Fact 2.1, we can find a left-separated sequence:
be a sequence of open sets which witnesses the left-separatedness of$M$. ThenBy the elementarity of , we may assume. Since::is left-separated inwitnesses that$ot(M\cap\delta)=\omega_{1},$has a left separated sequencewitnesses thatis not hereditarily separable, this contradicts to the secondof length , so. (2) follows from a similar argument.
countability of. Since. We see that $X$ has an open base of size
Now suppose
$X$
$X$
$X$
has a dense subset of size
has an open base of size
is , so is . Since. We see that. Now fix an open basewith$\{O_{i} : i<\gamma\}$ is a base for some. By the defimtion of the topology of. Since
is second countable,
is an open base for. On the other hand, since
which is a base forhas a countable subset. We see that, there issuch that
$X$$\{O_{i}: i<\gamma\}\in M,$
is an open base for . Suppose otherwise. Since
there is $x\in X\cap M$ and an open $O\in M$ such that there is no $i<\gamma$ with

Proof. First suppose to the contrary that$cf(\delta)<\delta$$f$$cf(\delta)arrow\delta$$M\cap cf(\delta)$$M\cap\delta$$cf(\delta)<\delta$$Y$$<\delta$$\langle O_{i}:i<\delta\rangle$$\langle x_{i}:i<\delta\rangle$$\langle x_{i}$$\langle x_{i}:i<\delta\rangle,$$\langle O_{i}\cap M$$i\in M\cap\delta\rangle$$\langle x_{i}$$\langle O_{i}:i<\delta\rangle\in M$$X_{M}$$i\in M\cap\delta\rangle$$X_{M}$$\langle x_{i}:i\in M\cap\delta\rangle$$\omega_{1}$$i<\delta\rangle.$$X_{M}$$X_{M}$$2^{<\delta}=\delta$$<\delta$$<\delta,$$T_{3}$$\leq 2^{<\delta}=\delta$$\mathcal{B}=\{O_{i} :i<\delta\}$$\mathcal{B}\in M$$X_{M},$$\gamma<\delta$$\mathcal{B}’$$\mathcal{B}_{M}$$ot(M\cap\delta)=\omega_{1}$$X_{M}$$X_{M}$$\mathcal{B}_{M}=\{O_{i}\cap M : i\in M\cap\delta\}$$\{O_{i} :$X_{M}$$\gamma\in M\cap\delta$$X_{M}$$\mathcal{B}’\subseteq\{O_{i}\cap M : i\in M\cap\gamma\}$i<\gamma\}$$x\in O_{i}\subseteq O.$

35, hence there is $i\in M\cap\gamma$ withby the elementarity of $M$ , this is a contradiction.

$x\in O_{i}\cap M\underline{\subset}O\cap M.$$\mathcal{B}’\subseteq\{O_{i}\cap M : i\in M\cap\gamma\}$Then$x\in O_{i}\subseteq O$$\square$

We use the following fact:

is completely $metrizable_{j}$ separable, and uncountFact 2.5 (Tall [3]). Suppose
oris uncountable, then $X_{M}=X.$
able.
$X_{M}$$If|X|\leq 2^{\omega}$$|\mathbb{R}\cap M|$andProposition 2.6. Suppose
for every oegular cardinal with$2^{\omega}<\kappa<|X|$ .
is completely metrizable, $separable_{f}$ and uncountable, thenIf$2^{<\kappa}=\kappa$$2^{\omega}>\omega_{1}$$\kappa$$X_{M}$$X=X_{M}.$, we are done by Fact 2.5. Suppose $|X|>2^{\omega}$ . Let $\kappa\in M$ beis regular uncountable by Lemma 2.4, and sincesuch that $ot(M A \kappa)=\omega_{1}.$$|M\cap|X||=|X_{M}|=2^{\omega}>\omega_{1}=|M\cap\kappa|$ , we have $\kappa<|X|.$by Lemma 2.4, butby our assumption. So, thenlf. Thus. Then
this is a contradiction. Thus we have$|M\cap 2^{\omega}|=|MA\mathbb{R}|$ is uncountable, so we have $X_{M}=X$ by Fact 2.5 again.

Proof. lf$|X|\leq 2^{\omega}$$\kappa$$\kappa>2^{\omega}$$2^{<\kappa}=\kappa$$|X|\leq 2^{<\kappa}=\kappa$$|M\cap 2^{\omega}|\geq|M\cap\kappa|=\omega_{1}$$\kappa\leq 2^{\omega}$$\square$Now we have the following conclusion.

and $2^{K}=\kappa^{+}for$ every cardinal(1) Suppose
Corollary 2.7.
is completely metrizable, $separable_{f}$ and uncountThen for every
ifable, then $X_{M}=X.$
is completely $metrizable_{f}$ separable,. For every $X$ , if(2) Supposeand uncountable, but $X_{M}\neq X$ , then $|X|>(2^{\omega})^{+}.$$\kappa\geq 2^{\omega}.$$2^{\omega}>\omega_{1}$$X_{f}$$X_{M}$$2^{\omega}>\omega_{\lambda}$$X_{M}$

This corollary shows that the aﬃrmative answer of an irrational problem is consistent with Martin’s Maximum: A standard argument shows that we can construct
for every, anda model of.ZFC in which Martin’s Maximum holds,$X$
is completely. In this model, we have that for every space , if
cardinal
metrizable , separable, and uncountable, then $X=X_{M}.$As mentioned before, Tall showed that if does not exist, thdn the aﬃrmative
answer of an irrational problem holds ([2]). The previous observation also showsexists (orthat the aﬃrmative answer of an irrational problem can hold even if
more strong large cardinal properties hold).
Under GCH, we also have the following:$2^{\omega}=\omega_{2}$$2^{\kappa}=\kappa^{-\succ}$$X_{M}$$\kappa\geq\omega_{2}$$\prime$

$0\#$

$o\#$

completely metrizable, separable,
Corollary 2.8. Suppose $GCH$. Supposeand uncountable. Let rc $\in M$ be such that $ot(M\cap\kappa)=\omega_{1}$ , Then $w(X)<|X|=\kappa.$
In $particular_{f}|X|$ must be regular.
$X_{M}\dot{u}$Proof. We haveagain.

$\kappa\leq|X|\leq 2^{<\kappa}=\kappa$by Lemma 2.4, and $w(X)<\kappa$ by Lemma 2.4$\square$36
REFERENCES[1] L. R. Junqueira, F. D. Tall, The topology of elementary submodels Topology. Appl. 82 (1998),239-266.
[2] F. D. Tall, If it looks and smells like the reals. Fund. Math. 163 $(20(0)1-11.$
F. D. Tall, An irrational problem. Fund. Math., 175 (2002) 259-269.
$|3]$(Usuba) ORGANIZATION OF ADVANCED SCIENCE AND TECHNOLOGY, KOBE UNIVERSITY,ROKKO-DAI 1-1, NADA, KOBE, 657-8501 JAPAN-maat address: usubaQpeople. $kobe-u$ . ac. jp$E$