Set theory problem about sets that contain at most one element
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It is problem 9 in A.Shen' book
Consider an equality whose left-hand side and right- hand side contain set variables and operations ∩, ∪, . Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.
enter image description here
discrete-mathematics elementary-set-theory
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
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add a comment |
$begingroup$
It is problem 9 in A.Shen' book
Consider an equality whose left-hand side and right- hand side contain set variables and operations ∩, ∪, . Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.
enter image description here
discrete-mathematics elementary-set-theory
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
$endgroup$
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Welcome to MSE. What were your attempts in solving this problem?
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– James
Jan 3 at 10:56
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Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19
add a comment |
$begingroup$
It is problem 9 in A.Shen' book
Consider an equality whose left-hand side and right- hand side contain set variables and operations ∩, ∪, . Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.
enter image description here
discrete-mathematics elementary-set-theory
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
$endgroup$
It is problem 9 in A.Shen' book
Consider an equality whose left-hand side and right- hand side contain set variables and operations ∩, ∪, . Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.
enter image description here
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
edited Jan 3 at 12:22
Unstandard Candle
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
asked Jan 3 at 10:37
Unstandard CandleUnstandard Candle
11
11
$begingroup$
Welcome to MSE. What were your attempts in solving this problem?
$endgroup$
– James
Jan 3 at 10:56
$begingroup$
Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19
add a comment |
$begingroup$
Welcome to MSE. What were your attempts in solving this problem?
$endgroup$
– James
Jan 3 at 10:56
$begingroup$
Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19
$begingroup$
Welcome to MSE. What were your attempts in solving this problem?
$endgroup$
– James
Jan 3 at 10:56
$begingroup$
Welcome to MSE. What were your attempts in solving this problem?
$endgroup$
– James
Jan 3 at 10:56
$begingroup$
Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19
$begingroup$
Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19
add a comment |
1 Answer
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That's interesting - seems like it sys something about $Bbb Z_2$-valued homomorphisms of Boolean algebras, sort of. Anyway, I bet the following works. First show this:
Lemma Fix $x$, and define $f(A)=Acap{x}$ for all $A$. Then $f(Acup B)=f(A)cup f(B)$, $f(Acap B)=f(A)cap f(B)$ and $f(Asetminus B)=f(A)setminus f(B)$ for all $A,B$.
Now say the two expressions are $phi(A_1,dots,A_n)$ and $psi(A_1,dots,A_n)$. Suppose $phi(A_1,dots,A_n)nepsi(A_1,dots,A_n)$. Then there exists $x$ with $xinphi(A_1,dots,A_n)$ and $xnotinpsi(A_1,dots,,A_n)$ (or vice versa). Hence $$phi((f(A_j))=f(phi((A_j)))ne f(psi((A_j)))=psi((f(A_j)).$$
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
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Thank you for this nice proof.
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– Unstandard Candle
Jan 4 at 2:22
add a comment |
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1 Answer
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$begingroup$
That's interesting - seems like it sys something about $Bbb Z_2$-valued homomorphisms of Boolean algebras, sort of. Anyway, I bet the following works. First show this:
Lemma Fix $x$, and define $f(A)=Acap{x}$ for all $A$. Then $f(Acup B)=f(A)cup f(B)$, $f(Acap B)=f(A)cap f(B)$ and $f(Asetminus B)=f(A)setminus f(B)$ for all $A,B$.
Now say the two expressions are $phi(A_1,dots,A_n)$ and $psi(A_1,dots,A_n)$. Suppose $phi(A_1,dots,A_n)nepsi(A_1,dots,A_n)$. Then there exists $x$ with $xinphi(A_1,dots,A_n)$ and $xnotinpsi(A_1,dots,,A_n)$ (or vice versa). Hence $$phi((f(A_j))=f(phi((A_j)))ne f(psi((A_j)))=psi((f(A_j)).$$
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
add a comment |
$begingroup$
That's interesting - seems like it sys something about $Bbb Z_2$-valued homomorphisms of Boolean algebras, sort of. Anyway, I bet the following works. First show this:
Lemma Fix $x$, and define $f(A)=Acap{x}$ for all $A$. Then $f(Acup B)=f(A)cup f(B)$, $f(Acap B)=f(A)cap f(B)$ and $f(Asetminus B)=f(A)setminus f(B)$ for all $A,B$.
Now say the two expressions are $phi(A_1,dots,A_n)$ and $psi(A_1,dots,A_n)$. Suppose $phi(A_1,dots,A_n)nepsi(A_1,dots,A_n)$. Then there exists $x$ with $xinphi(A_1,dots,A_n)$ and $xnotinpsi(A_1,dots,,A_n)$ (or vice versa). Hence $$phi((f(A_j))=f(phi((A_j)))ne f(psi((A_j)))=psi((f(A_j)).$$
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
add a comment |
$begingroup$
That's interesting - seems like it sys something about $Bbb Z_2$-valued homomorphisms of Boolean algebras, sort of. Anyway, I bet the following works. First show this:
Lemma Fix $x$, and define $f(A)=Acap{x}$ for all $A$. Then $f(Acup B)=f(A)cup f(B)$, $f(Acap B)=f(A)cap f(B)$ and $f(Asetminus B)=f(A)setminus f(B)$ for all $A,B$.
Now say the two expressions are $phi(A_1,dots,A_n)$ and $psi(A_1,dots,A_n)$. Suppose $phi(A_1,dots,A_n)nepsi(A_1,dots,A_n)$. Then there exists $x$ with $xinphi(A_1,dots,A_n)$ and $xnotinpsi(A_1,dots,,A_n)$ (or vice versa). Hence $$phi((f(A_j))=f(phi((A_j)))ne f(psi((A_j)))=psi((f(A_j)).$$
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
That's interesting - seems like it sys something about $Bbb Z_2$-valued homomorphisms of Boolean algebras, sort of. Anyway, I bet the following works. First show this:
Lemma Fix $x$, and define $f(A)=Acap{x}$ for all $A$. Then $f(Acup B)=f(A)cup f(B)$, $f(Acap B)=f(A)cap f(B)$ and $f(Asetminus B)=f(A)setminus f(B)$ for all $A,B$.
Now say the two expressions are $phi(A_1,dots,A_n)$ and $psi(A_1,dots,A_n)$. Suppose $phi(A_1,dots,A_n)nepsi(A_1,dots,A_n)$. Then there exists $x$ with $xinphi(A_1,dots,A_n)$ and $xnotinpsi(A_1,dots,,A_n)$ (or vice versa). Hence $$phi((f(A_j))=f(phi((A_j)))ne f(psi((A_j)))=psi((f(A_j)).$$
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
edited Jan 3 at 20:10
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 3 at 14:52
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
add a comment |
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
$begingroup$
Thank you for this nice proof.
$endgroup$
– Unstandard Candle
Jan 4 at 2:22
add a comment |
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$begingroup$
Welcome to MSE. What were your attempts in solving this problem?
$endgroup$
– James
Jan 3 at 10:56
$begingroup$
Just encounter it when reading the book. I guess it is an easy problem but cannot figure out the proof.
$endgroup$
– Unstandard Candle
Jan 3 at 12:19