The product of perfectly normal compact Hausdorff spaces is perfectly normal

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Question: Is it true that if $X$ is a compact Hausdorff space that is perfectly normal, then $X times X$ is also perfectly normal?

Recall that a topological space $X$ is called perfectly normal if and only if $X$ is normal and every closed set $A subseteq X$ is a $G_delta$-set ($A$ is a $G_delta$-set if and only if it is the intersection of countably many open sets of $X$).

Let $X$ be a compact Hausdorff space. In particular, this means that $X$ is compact. Consequently, $X times X$ is compact and Hausdorff. Thus, to show that $X times X$ is perfectly normal, it is enough to show that each closed set in $X times X$ is a $G_delta$-set. Thus, my question is really the following:

Question*: Is it true that If $X$ is a compact Hausdorff and perfectly normal space, then every closed set in $X times X$ is a $G_delta$-set?

If this turns out to be false, that's too bad. However, if this turns out to be true, then I have $2$ follow-up questions, where we get rid of some assumptions regarding compactness or Hausdorffness.

Bonus question 1: If $X$ is a topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

Bonus question 2: If $X$ is a Hausdorff topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

As always, any help with any of the above questions will be greatly appreciated.

general-topology compactness

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Question: Is it true that if $X$ is a compact Hausdorff space that is perfectly normal, then $X times X$ is also perfectly normal?

Recall that a topological space $X$ is called perfectly normal if and only if $X$ is normal and every closed set $A subseteq X$ is a $G_delta$-set ($A$ is a $G_delta$-set if and only if it is the intersection of countably many open sets of $X$).

Let $X$ be a compact Hausdorff space. In particular, this means that $X$ is compact. Consequently, $X times X$ is compact and Hausdorff. Thus, to show that $X times X$ is perfectly normal, it is enough to show that each closed set in $X times X$ is a $G_delta$-set. Thus, my question is really the following:

Question*: Is it true that If $X$ is a compact Hausdorff and perfectly normal space, then every closed set in $X times X$ is a $G_delta$-set?

If this turns out to be false, that's too bad. However, if this turns out to be true, then I have $2$ follow-up questions, where we get rid of some assumptions regarding compactness or Hausdorffness.

Bonus question 1: If $X$ is a topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

Bonus question 2: If $X$ is a Hausdorff topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

As always, any help with any of the above questions will be greatly appreciated.

general-topology compactness

share|cite|improve this question

edited 2 days ago

asked 2 days ago

Pawel

2,780820

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Question: Is it true that if $X$ is a compact Hausdorff space that is perfectly normal, then $X times X$ is also perfectly normal?

Recall that a topological space $X$ is called perfectly normal if and only if $X$ is normal and every closed set $A subseteq X$ is a $G_delta$-set ($A$ is a $G_delta$-set if and only if it is the intersection of countably many open sets of $X$).

Let $X$ be a compact Hausdorff space. In particular, this means that $X$ is compact. Consequently, $X times X$ is compact and Hausdorff. Thus, to show that $X times X$ is perfectly normal, it is enough to show that each closed set in $X times X$ is a $G_delta$-set. Thus, my question is really the following:

Question*: Is it true that If $X$ is a compact Hausdorff and perfectly normal space, then every closed set in $X times X$ is a $G_delta$-set?

If this turns out to be false, that's too bad. However, if this turns out to be true, then I have $2$ follow-up questions, where we get rid of some assumptions regarding compactness or Hausdorffness.

Bonus question 1: If $X$ is a topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

Bonus question 2: If $X$ is a Hausdorff topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

As always, any help with any of the above questions will be greatly appreciated.

general-topology compactness

share|cite|improve this question

edited 2 days ago

asked 2 days ago

Pawel

2,780820

Question: Is it true that if $X$ is a compact Hausdorff space that is perfectly normal, then $X times X$ is also perfectly normal?

Recall that a topological space $X$ is called perfectly normal if and only if $X$ is normal and every closed set $A subseteq X$ is a $G_delta$-set ($A$ is a $G_delta$-set if and only if it is the intersection of countably many open sets of $X$).

Let $X$ be a compact Hausdorff space. In particular, this means that $X$ is compact. Consequently, $X times X$ is compact and Hausdorff. Thus, to show that $X times X$ is perfectly normal, it is enough to show that each closed set in $X times X$ is a $G_delta$-set. Thus, my question is really the following:

Question*: Is it true that If $X$ is a compact Hausdorff and perfectly normal space, then every closed set in $X times X$ is a $G_delta$-set?

If this turns out to be false, that's too bad. However, if this turns out to be true, then I have $2$ follow-up questions, where we get rid of some assumptions regarding compactness or Hausdorffness.

Bonus question 1: If $X$ is a topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

Bonus question 2: If $X$ is a Hausdorff topological space such that every closed set is a $G_delta$-set, then every closed set in $X times X$ is a $G_delta$-set.

As always, any help with any of the above questions will be greatly appreciated.

general-topology compactness

share|cite|improve this question

edited 2 days ago

asked 2 days ago

Pawel

2,780820

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share|cite|improve this question

edited 2 days ago

edited 2 days ago

edited 2 days ago

asked 2 days ago

Pawel

2,780820

asked 2 days ago

Pawel

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asked 2 days ago

Pawel

2,780820

2,780820

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1 Answer
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The answer is "no". This follows from Corollary 2 of

Katětov, Miroslav. "Complete normality of Cartesian products." Fundamenta Mathematicae 35.1 (1948): 271-274.

See http://matwbn.icm.edu.pl/ksiazki/fm/fm35/fm35125.pdf.

Assume it were true that for any perfectly normal compact Hausdorff space $X$ also $X times X$ is perfectly normal. Then also $X times X times X times X$ would be perfectly normal, hence also $Y = X times X times X$ because it embeds into $X times X times X times X$. Since perfectly normal spaces are heritarily normal (which is denoted as "completely normal" in the above paper), Corollary 2 implies that $X$ is metrizable.

In other words, if $mathcalP$ is a class of perfectly normal compact Hausdorff spaces such that

$$(ast) phantomx X in mathcalP Rightarrow X times X in mathcalP$$

then $mathcalP$ is contained in the class $mathcalCM$ of compact metrizable spaces. In fact, $mathcalCM$ is the biggest class having property $(ast)$.

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Paul Frost

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
2
down vote



accepted

The answer is "no". This follows from Corollary 2 of

Katětov, Miroslav. "Complete normality of Cartesian products." Fundamenta Mathematicae 35.1 (1948): 271-274.

See http://matwbn.icm.edu.pl/ksiazki/fm/fm35/fm35125.pdf.

Assume it were true that for any perfectly normal compact Hausdorff space $X$ also $X times X$ is perfectly normal. Then also $X times X times X times X$ would be perfectly normal, hence also $Y = X times X times X$ because it embeds into $X times X times X times X$. Since perfectly normal spaces are heritarily normal (which is denoted as "completely normal" in the above paper), Corollary 2 implies that $X$ is metrizable.

In other words, if $mathcalP$ is a class of perfectly normal compact Hausdorff spaces such that

$$(ast) phantomx X in mathcalP Rightarrow X times X in mathcalP$$

then $mathcalP$ is contained in the class $mathcalCM$ of compact metrizable spaces. In fact, $mathcalCM$ is the biggest class having property $(ast)$.

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Paul Frost

3,393320

add a comment | 

up vote
2
down vote



accepted

The answer is "no". This follows from Corollary 2 of

Katětov, Miroslav. "Complete normality of Cartesian products." Fundamenta Mathematicae 35.1 (1948): 271-274.

See http://matwbn.icm.edu.pl/ksiazki/fm/fm35/fm35125.pdf.

Assume it were true that for any perfectly normal compact Hausdorff space $X$ also $X times X$ is perfectly normal. Then also $X times X times X times X$ would be perfectly normal, hence also $Y = X times X times X$ because it embeds into $X times X times X times X$. Since perfectly normal spaces are heritarily normal (which is denoted as "completely normal" in the above paper), Corollary 2 implies that $X$ is metrizable.

In other words, if $mathcalP$ is a class of perfectly normal compact Hausdorff spaces such that

$$(ast) phantomx X in mathcalP Rightarrow X times X in mathcalP$$

then $mathcalP$ is contained in the class $mathcalCM$ of compact metrizable spaces. In fact, $mathcalCM$ is the biggest class having property $(ast)$.

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Paul Frost

3,393320

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The answer is "no". This follows from Corollary 2 of

Katětov, Miroslav. "Complete normality of Cartesian products." Fundamenta Mathematicae 35.1 (1948): 271-274.

See http://matwbn.icm.edu.pl/ksiazki/fm/fm35/fm35125.pdf.

Assume it were true that for any perfectly normal compact Hausdorff space $X$ also $X times X$ is perfectly normal. Then also $X times X times X times X$ would be perfectly normal, hence also $Y = X times X times X$ because it embeds into $X times X times X times X$. Since perfectly normal spaces are heritarily normal (which is denoted as "completely normal" in the above paper), Corollary 2 implies that $X$ is metrizable.

In other words, if $mathcalP$ is a class of perfectly normal compact Hausdorff spaces such that

$$(ast) phantomx X in mathcalP Rightarrow X times X in mathcalP$$

then $mathcalP$ is contained in the class $mathcalCM$ of compact metrizable spaces. In fact, $mathcalCM$ is the biggest class having property $(ast)$.

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Paul Frost

3,393320

The answer is "no". This follows from Corollary 2 of

Katětov, Miroslav. "Complete normality of Cartesian products." Fundamenta Mathematicae 35.1 (1948): 271-274.

See http://matwbn.icm.edu.pl/ksiazki/fm/fm35/fm35125.pdf.

Assume it were true that for any perfectly normal compact Hausdorff space $X$ also $X times X$ is perfectly normal. Then also $X times X times X times X$ would be perfectly normal, hence also $Y = X times X times X$ because it embeds into $X times X times X times X$. Since perfectly normal spaces are heritarily normal (which is denoted as "completely normal" in the above paper), Corollary 2 implies that $X$ is metrizable.

In other words, if $mathcalP$ is a class of perfectly normal compact Hausdorff spaces such that

$$(ast) phantomx X in mathcalP Rightarrow X times X in mathcalP$$

then $mathcalP$ is contained in the class $mathcalCM$ of compact metrizable spaces. In fact, $mathcalCM$ is the biggest class having property $(ast)$.

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Paul Frost

3,393320

share|cite|improve this answer

share|cite|improve this answer

edited 20 hours ago

edited 20 hours ago

edited 20 hours ago

answered yesterday

Paul Frost

3,393320

answered yesterday

Paul Frost

3,393320

answered yesterday

Paul Frost

3,393320

3,393320

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