Is there a way to prove the Banach-Tarski theorem from just the Partition Principle?
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The Banach-Tarksi theorem is one of the most notorious "counterintuitive consequences" of choice (and in fact in all of mathematics), making heavy use of non-measurable sets. However, full choice is not needed, and it is well-known that under $ZF$ the Hahn-Banach theorem suffices, which in turn is implied by the ultrafilter lemma. In many cases, it turns out the "pathological" and "counterintuitive" consequences of choice actually follow from much less.
The Partition Principle (implied by $AC$) and even the Weak Partition Principle (if $P$ is a quotient of $S$, $|P| ngtr |S|$) already imply non-measurable sets. $AC$ implies $PP$ but whether $PP$ implies $AC$ is a long-open problem, so whether this is a strictly weaker principle is unknown, but these statements are to many people so intuitive and unobjectionable that their failure, e.g., "a set can be partitioned into strictly more equivalence classes than it has elements" seems to many people much "worse" than Banach-Tarski. From the Partition Principle or any similar principle is there any known way to prove Banach-Tarski, or a similar result? Is there any known way to get from PP to the Ultrafilter lemma?
set-theory
asked Sep 10 at 18:10
penguin_surprise
1106
add a comment |Â
up vote
4
down vote
favorite
The Banach-Tarksi theorem is one of the most notorious "counterintuitive consequences" of choice (and in fact in all of mathematics), making heavy use of non-measurable sets. However, full choice is not needed, and it is well-known that under $ZF$ the Hahn-Banach theorem suffices, which in turn is implied by the ultrafilter lemma. In many cases, it turns out the "pathological" and "counterintuitive" consequences of choice actually follow from much less.
The Partition Principle (implied by $AC$) and even the Weak Partition Principle (if $P$ is a quotient of $S$, $|P| ngtr |S|$) already imply non-measurable sets. $AC$ implies $PP$ but whether $PP$ implies $AC$ is a long-open problem, so whether this is a strictly weaker principle is unknown, but these statements are to many people so intuitive and unobjectionable that their failure, e.g., "a set can be partitioned into strictly more equivalence classes than it has elements" seems to many people much "worse" than Banach-Tarski. From the Partition Principle or any similar principle is there any known way to prove Banach-Tarski, or a similar result? Is there any known way to get from PP to the Ultrafilter lemma?
set-theory
asked Sep 10 at 18:10
penguin_surprise
1106
If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
3
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The Banach-Tarksi theorem is one of the most notorious "counterintuitive consequences" of choice (and in fact in all of mathematics), making heavy use of non-measurable sets. However, full choice is not needed, and it is well-known that under $ZF$ the Hahn-Banach theorem suffices, which in turn is implied by the ultrafilter lemma. In many cases, it turns out the "pathological" and "counterintuitive" consequences of choice actually follow from much less.
The Partition Principle (implied by $AC$) and even the Weak Partition Principle (if $P$ is a quotient of $S$, $|P| ngtr |S|$) already imply non-measurable sets. $AC$ implies $PP$ but whether $PP$ implies $AC$ is a long-open problem, so whether this is a strictly weaker principle is unknown, but these statements are to many people so intuitive and unobjectionable that their failure, e.g., "a set can be partitioned into strictly more equivalence classes than it has elements" seems to many people much "worse" than Banach-Tarski. From the Partition Principle or any similar principle is there any known way to prove Banach-Tarski, or a similar result? Is there any known way to get from PP to the Ultrafilter lemma?
set-theory
asked Sep 10 at 18:10
penguin_surprise
1106
The Banach-Tarksi theorem is one of the most notorious "counterintuitive consequences" of choice (and in fact in all of mathematics), making heavy use of non-measurable sets. However, full choice is not needed, and it is well-known that under $ZF$ the Hahn-Banach theorem suffices, which in turn is implied by the ultrafilter lemma. In many cases, it turns out the "pathological" and "counterintuitive" consequences of choice actually follow from much less.
The Partition Principle (implied by $AC$) and even the Weak Partition Principle (if $P$ is a quotient of $S$, $|P| ngtr |S|$) already imply non-measurable sets. $AC$ implies $PP$ but whether $PP$ implies $AC$ is a long-open problem, so whether this is a strictly weaker principle is unknown, but these statements are to many people so intuitive and unobjectionable that their failure, e.g., "a set can be partitioned into strictly more equivalence classes than it has elements" seems to many people much "worse" than Banach-Tarski. From the Partition Principle or any similar principle is there any known way to prove Banach-Tarski, or a similar result? Is there any known way to get from PP to the Ultrafilter lemma?
set-theory
set-theory
asked Sep 10 at 18:10
penguin_surprise
1106
asked Sep 10 at 18:10
penguin_surprise
1106
asked Sep 10 at 18:10
penguin_surprise
1106
asked Sep 10 at 18:10
penguin_surprise
1106
asked Sep 10 at 18:10
penguin_surprise
1106
1106
If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
3
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33
add a comment |Â
If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
3
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33
If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
3
3
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33
add a comment |Â
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If there was a way to get $mathsfPP$ from the ultrafilter lemma, the answer whether $mathsfPPRightarrowmathsfAC$ would have to be negative, since the ultrafilter lemma is known to be strictly weaker than Choice. So the answer to your last question must be "no."
– Malice Vidrine
Sep 10 at 20:20
Not as far as I know.
– Asaf Karagila♦
Sep 10 at 22:15
3
@Malice: BPI definitely cannot imply PP, since in Cohen's model BPI holds but PP fails (since DC fails, and PP implies DC).
– Asaf Karagila♦
Sep 11 at 5:33