Validating a proof for necessary and sufficient conditions on sets A & B such that $P(A) cup P(B) = P(Acup B)$ The Next CEO of Stack OverflowProof on disjoint union of sets $A$ and $B$Am I correct? State the necessary and sufficient condition for R to be an equivalence relation on A.Difference between “necessary” and “necessary but not sufficient”?Show that a necessary and sufficient condition that $(Acap B)cup C =Acap(Bcup C)$ is that $Csubseteq A$.Sufficient and necessary conditions for |A $circ$ B| = |A $times$ B|Necessary and sufficient conditions $(L_1 circ L_2)/L_2 = L_1$Which condition is necessary and sufficient for $f^-1(f(C)) = C$Proof by cases for setsSufficient and necessary conditions for order relations describing a certain system of setsA sufficient and necessary condition for $Acup (B cap C) = (A cup B) cap C$
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Validating a proof for necessary and sufficient conditions on sets A & B such that $P(A) cup P(B) = P(Acup B)$
The Next CEO of Stack OverflowProof on disjoint union of sets $A$ and $B$Am I correct? State the necessary and sufficient condition for R to be an equivalence relation on A.Difference between “necessary” and “necessary but not sufficient”?Show that a necessary and sufficient condition that $(Acap B)cup C =Acap(Bcup C)$ is that $Csubseteq A$.Sufficient and necessary conditions for |A $circ$ B| = |A $times$ B|Necessary and sufficient conditions $(L_1 circ L_2)/L_2 = L_1$Which condition is necessary and sufficient for $f^-1(f(C)) = C$Proof by cases for setsSufficient and necessary conditions for order relations describing a certain system of setsA sufficient and necessary condition for $Acup (B cap C) = (A cup B) cap C$
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Is the solution correct for this?
Problem: Find necessary and sufficient conditions on sets $A$ and $B$ such that the following relation holds among the power sets: $P(A)cup P(B) = P(Acup B)$.
Answer: Suppose first that $Ansubseteq B$ and $Bnsubseteq A$. Then there exists $a in A setminus B$ and $b in B setminus A$. The subset, a,b of $A cup B$ is not a subset of $A$, nor is it a subset of $B$. Thus, we do not have equality.
Suppose that $Asubseteq B$, then $A cup B = B$. So, $P(B) subseteq P(A) cup P(B)$ by the definition of power sets. Moreover, all subsets of $A$ are subset of $B$, thus you arrive to $P(A) cup P(B) = P(A cup B)$. It is the same if $B subseteq A$.
Thus, the condition is that $A subseteq B$ or $Bsubseteq A.$
discrete-mathematics proof-verification elementary-set-theory
asked 2 days ago
brucemcmcbrucemcmc
396
$endgroup$
add a comment |
$begingroup$
Is the solution correct for this?
Problem: Find necessary and sufficient conditions on sets $A$ and $B$ such that the following relation holds among the power sets: $P(A)cup P(B) = P(Acup B)$.
Answer: Suppose first that $Ansubseteq B$ and $Bnsubseteq A$. Then there exists $a in A setminus B$ and $b in B setminus A$. The subset, a,b of $A cup B$ is not a subset of $A$, nor is it a subset of $B$. Thus, we do not have equality.
Suppose that $Asubseteq B$, then $A cup B = B$. So, $P(B) subseteq P(A) cup P(B)$ by the definition of power sets. Moreover, all subsets of $A$ are subset of $B$, thus you arrive to $P(A) cup P(B) = P(A cup B)$. It is the same if $B subseteq A$.
Thus, the condition is that $A subseteq B$ or $Bsubseteq A.$
discrete-mathematics proof-verification elementary-set-theory
asked 2 days ago
brucemcmcbrucemcmc
396
$endgroup$
$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago
add a comment |
$begingroup$
Is the solution correct for this?
Problem: Find necessary and sufficient conditions on sets $A$ and $B$ such that the following relation holds among the power sets: $P(A)cup P(B) = P(Acup B)$.
Answer: Suppose first that $Ansubseteq B$ and $Bnsubseteq A$. Then there exists $a in A setminus B$ and $b in B setminus A$. The subset, a,b of $A cup B$ is not a subset of $A$, nor is it a subset of $B$. Thus, we do not have equality.
Suppose that $Asubseteq B$, then $A cup B = B$. So, $P(B) subseteq P(A) cup P(B)$ by the definition of power sets. Moreover, all subsets of $A$ are subset of $B$, thus you arrive to $P(A) cup P(B) = P(A cup B)$. It is the same if $B subseteq A$.
Thus, the condition is that $A subseteq B$ or $Bsubseteq A.$
discrete-mathematics proof-verification elementary-set-theory
asked 2 days ago
brucemcmcbrucemcmc
396
$endgroup$
Is the solution correct for this?
Problem: Find necessary and sufficient conditions on sets $A$ and $B$ such that the following relation holds among the power sets: $P(A)cup P(B) = P(Acup B)$.
Answer: Suppose first that $Ansubseteq B$ and $Bnsubseteq A$. Then there exists $a in A setminus B$ and $b in B setminus A$. The subset, a,b of $A cup B$ is not a subset of $A$, nor is it a subset of $B$. Thus, we do not have equality.
Suppose that $Asubseteq B$, then $A cup B = B$. So, $P(B) subseteq P(A) cup P(B)$ by the definition of power sets. Moreover, all subsets of $A$ are subset of $B$, thus you arrive to $P(A) cup P(B) = P(A cup B)$. It is the same if $B subseteq A$.
Thus, the condition is that $A subseteq B$ or $Bsubseteq A.$
discrete-mathematics proof-verification elementary-set-theory
discrete-mathematics proof-verification elementary-set-theory
asked 2 days ago
brucemcmcbrucemcmc
396
asked 2 days ago
brucemcmcbrucemcmc
396
edited 2 days ago
brucemcmc
asked 2 days ago
brucemcmcbrucemcmc
396
asked 2 days ago
brucemcmcbrucemcmc
396
asked 2 days ago
brucemcmcbrucemcmc
396
396
$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago
add a comment |
$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago
$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago
$begingroup$
@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The first part is good.
When you say
Therefore, $P(B) subseteq P(A) cup P(B).$
there is nothing "therefore" about it. $Xsubseteq Ycup X$ is basically the definition of $cup$.
I would prefer to see some justification for why
$$
Asubseteq Bimplies P(A)subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
answered 2 days ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
add a comment |
$begingroup$
Solution: $A subseteq (Acup B) implies P(A) subseteq P(Acup B) ; B subseteq (A cup B) implies P(B) subseteq P(A cup B)$.
So, $P(A) cup P(B) subseteq P(Acup B)$ always.
Conversely, lets claim that $P(Acup B) subseteq P(A) cup P(B)$ if and only if $Asubseteq B$ or $Bsubseteq A$.
Suppose $P(Acup B) subseteq P(A) cup P(B)$.
If $Ansubseteq B$ and $ B nsubseteq A$. Then there exists some $xin A$ such that $x notin B$ and $yin B$ such that $ynotin A$. Let S= x,y
Then it is clear that $Ssubseteq (Acup B) implies Sin P(Acup B) subseteq P(A) cup P(B)$.
Also, $Snsubseteq A$ and $S nsubseteq B$ therefore, $xin S$, but $xnotin B$ and $yin S$, but $ynotin A$. Hence $Snotin P(A)$ and $S notin P(B) implies Snotin P(A) cup P(B)$ which is a contradiction because $Sin P(Acup B) subseteq P(A) cup P(B)$.
So, $Asubseteq B$ or $Bsubseteq A$.
If $Asubseteq B$ or $Bsubseteq A implies Acup Bsubseteq B$ or $Acup Bsubset A$.
So, $P(Acup B) subseteq P(B)$ or $P(Acup B) subseteq P(A) implies P(Acup B) subseteq P(A) cup P(B)$.
Thus, $P(Acup B) = P(A)cup P(B)$ if and only if $A subseteq B$ or $Bsubseteq A$.
answered 2 days ago
bengraham12bengraham12
194
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first part is good.
When you say
Therefore, $P(B) subseteq P(A) cup P(B).$
there is nothing "therefore" about it. $Xsubseteq Ycup X$ is basically the definition of $cup$.
I would prefer to see some justification for why
$$
Asubseteq Bimplies P(A)subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
answered 2 days ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
add a comment |
$begingroup$
The first part is good.
When you say
Therefore, $P(B) subseteq P(A) cup P(B).$
there is nothing "therefore" about it. $Xsubseteq Ycup X$ is basically the definition of $cup$.
I would prefer to see some justification for why
$$
Asubseteq Bimplies P(A)subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
answered 2 days ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
add a comment |
$begingroup$
The first part is good.
When you say
Therefore, $P(B) subseteq P(A) cup P(B).$
there is nothing "therefore" about it. $Xsubseteq Ycup X$ is basically the definition of $cup$.
I would prefer to see some justification for why
$$
Asubseteq Bimplies P(A)subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
answered 2 days ago
ArthurArthur
120k7121206
$endgroup$
The first part is good.
When you say
Therefore, $P(B) subseteq P(A) cup P(B).$
there is nothing "therefore" about it. $Xsubseteq Ycup X$ is basically the definition of $cup$.
I would prefer to see some justification for why
$$
Asubseteq Bimplies P(A)subseteq P(B)
$$
It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.
answered 2 days ago
ArthurArthur
120k7121206
answered 2 days ago
ArthurArthur
120k7121206
answered 2 days ago
ArthurArthur
120k7121206
answered 2 days ago
ArthurArthur
120k7121206
120k7121206
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
add a comment |
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
by definition of power set can we not say that $A subseteq B implies P(A) subseteq P(B)$ ? Also, instead of therefore, can i just say "So." Moreover thank you for the response.
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
Do you mean $Asubseteq Bimplies P(A)le P(B)?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
@AdrianKeister These aren't probabilities, but power sets.
$endgroup$
– Arthur
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
Oh, gotcha; thanks!
$endgroup$
– Adrian Keister
2 days ago
add a comment |
$begingroup$
Solution: $A subseteq (Acup B) implies P(A) subseteq P(Acup B) ; B subseteq (A cup B) implies P(B) subseteq P(A cup B)$.
So, $P(A) cup P(B) subseteq P(Acup B)$ always.
Conversely, lets claim that $P(Acup B) subseteq P(A) cup P(B)$ if and only if $Asubseteq B$ or $Bsubseteq A$.
Suppose $P(Acup B) subseteq P(A) cup P(B)$.
If $Ansubseteq B$ and $ B nsubseteq A$. Then there exists some $xin A$ such that $x notin B$ and $yin B$ such that $ynotin A$. Let S= x,y
Then it is clear that $Ssubseteq (Acup B) implies Sin P(Acup B) subseteq P(A) cup P(B)$.
Also, $Snsubseteq A$ and $S nsubseteq B$ therefore, $xin S$, but $xnotin B$ and $yin S$, but $ynotin A$. Hence $Snotin P(A)$ and $S notin P(B) implies Snotin P(A) cup P(B)$ which is a contradiction because $Sin P(Acup B) subseteq P(A) cup P(B)$.
So, $Asubseteq B$ or $Bsubseteq A$.
If $Asubseteq B$ or $Bsubseteq A implies Acup Bsubseteq B$ or $Acup Bsubset A$.
So, $P(Acup B) subseteq P(B)$ or $P(Acup B) subseteq P(A) implies P(Acup B) subseteq P(A) cup P(B)$.
Thus, $P(Acup B) = P(A)cup P(B)$ if and only if $A subseteq B$ or $Bsubseteq A$.
answered 2 days ago
bengraham12bengraham12
194
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
add a comment |
$begingroup$
Solution: $A subseteq (Acup B) implies P(A) subseteq P(Acup B) ; B subseteq (A cup B) implies P(B) subseteq P(A cup B)$.
So, $P(A) cup P(B) subseteq P(Acup B)$ always.
Conversely, lets claim that $P(Acup B) subseteq P(A) cup P(B)$ if and only if $Asubseteq B$ or $Bsubseteq A$.
Suppose $P(Acup B) subseteq P(A) cup P(B)$.
If $Ansubseteq B$ and $ B nsubseteq A$. Then there exists some $xin A$ such that $x notin B$ and $yin B$ such that $ynotin A$. Let S= x,y
Then it is clear that $Ssubseteq (Acup B) implies Sin P(Acup B) subseteq P(A) cup P(B)$.
Also, $Snsubseteq A$ and $S nsubseteq B$ therefore, $xin S$, but $xnotin B$ and $yin S$, but $ynotin A$. Hence $Snotin P(A)$ and $S notin P(B) implies Snotin P(A) cup P(B)$ which is a contradiction because $Sin P(Acup B) subseteq P(A) cup P(B)$.
So, $Asubseteq B$ or $Bsubseteq A$.
If $Asubseteq B$ or $Bsubseteq A implies Acup Bsubseteq B$ or $Acup Bsubset A$.
So, $P(Acup B) subseteq P(B)$ or $P(Acup B) subseteq P(A) implies P(Acup B) subseteq P(A) cup P(B)$.
Thus, $P(Acup B) = P(A)cup P(B)$ if and only if $A subseteq B$ or $Bsubseteq A$.
answered 2 days ago
bengraham12bengraham12
194
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
add a comment |
$begingroup$
Solution: $A subseteq (Acup B) implies P(A) subseteq P(Acup B) ; B subseteq (A cup B) implies P(B) subseteq P(A cup B)$.
So, $P(A) cup P(B) subseteq P(Acup B)$ always.
Conversely, lets claim that $P(Acup B) subseteq P(A) cup P(B)$ if and only if $Asubseteq B$ or $Bsubseteq A$.
Suppose $P(Acup B) subseteq P(A) cup P(B)$.
If $Ansubseteq B$ and $ B nsubseteq A$. Then there exists some $xin A$ such that $x notin B$ and $yin B$ such that $ynotin A$. Let S= x,y
Then it is clear that $Ssubseteq (Acup B) implies Sin P(Acup B) subseteq P(A) cup P(B)$.
Also, $Snsubseteq A$ and $S nsubseteq B$ therefore, $xin S$, but $xnotin B$ and $yin S$, but $ynotin A$. Hence $Snotin P(A)$ and $S notin P(B) implies Snotin P(A) cup P(B)$ which is a contradiction because $Sin P(Acup B) subseteq P(A) cup P(B)$.
So, $Asubseteq B$ or $Bsubseteq A$.
If $Asubseteq B$ or $Bsubseteq A implies Acup Bsubseteq B$ or $Acup Bsubset A$.
So, $P(Acup B) subseteq P(B)$ or $P(Acup B) subseteq P(A) implies P(Acup B) subseteq P(A) cup P(B)$.
Thus, $P(Acup B) = P(A)cup P(B)$ if and only if $A subseteq B$ or $Bsubseteq A$.
answered 2 days ago
bengraham12bengraham12
194
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$endgroup$
Solution: $A subseteq (Acup B) implies P(A) subseteq P(Acup B) ; B subseteq (A cup B) implies P(B) subseteq P(A cup B)$.
So, $P(A) cup P(B) subseteq P(Acup B)$ always.
Conversely, lets claim that $P(Acup B) subseteq P(A) cup P(B)$ if and only if $Asubseteq B$ or $Bsubseteq A$.
Suppose $P(Acup B) subseteq P(A) cup P(B)$.
If $Ansubseteq B$ and $ B nsubseteq A$. Then there exists some $xin A$ such that $x notin B$ and $yin B$ such that $ynotin A$. Let S= x,y
Then it is clear that $Ssubseteq (Acup B) implies Sin P(Acup B) subseteq P(A) cup P(B)$.
Also, $Snsubseteq A$ and $S nsubseteq B$ therefore, $xin S$, but $xnotin B$ and $yin S$, but $ynotin A$. Hence $Snotin P(A)$ and $S notin P(B) implies Snotin P(A) cup P(B)$ which is a contradiction because $Sin P(Acup B) subseteq P(A) cup P(B)$.
So, $Asubseteq B$ or $Bsubseteq A$.
If $Asubseteq B$ or $Bsubseteq A implies Acup Bsubseteq B$ or $Acup Bsubset A$.
So, $P(Acup B) subseteq P(B)$ or $P(Acup B) subseteq P(A) implies P(Acup B) subseteq P(A) cup P(B)$.
Thus, $P(Acup B) = P(A)cup P(B)$ if and only if $A subseteq B$ or $Bsubseteq A$.
answered 2 days ago
bengraham12bengraham12
194
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bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
bengraham12bengraham12
194
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
bengraham12bengraham12
194
answered 2 days ago
bengraham12bengraham12
194
194
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
bengraham12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
add a comment |
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Thank you! @bengraham12
$endgroup$
– brucemcmc
2 days ago
add a comment |
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$begingroup$
You need to clean up your notation. Sets can admit the symbols $subseteq$ and $cap$ and $cup,$ but probabilities do not - they are real numbers. For example, $P(B)subseteq P(A)cup P(B)$ is nonsense.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
we are dealing with power sets. My bad that I did not make that clear. @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
$begingroup$
Gotcha, thanks!
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
No worries, my bad for the confusion! @AdrianKeister
$endgroup$
– brucemcmc
2 days ago
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@AdrianKeister its powerset, not probability.
$endgroup$
– Bertrand Wittgenstein's Ghost
2 days ago