Proof of a formula for the power set of a union of two setsProve $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin...
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Proof of a formula for the power set of a union of two sets
Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.Set theory proofUnion closure of a set of five finite setsStep-by-step help using the distributive law in set theoryUnion of sets proofProof that union of powersets equals powerset of union of two sets if one is subset of anotherPower sets and subsets proof help?Unary union proofProving Union and Intersection of Power SetsHow to calculate expression of union of sets?How to prove the equation for the union of two sets?
$begingroup$
I need to prove that:
$2^{Acup B} =left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$
Note that $2^{A} = P(A)$ (The power set of A).
While I can prove the Right - Left direction fairly easily, It seems very hard for me to prove the Left - Right direction. Can I get some help with that?
How do I prove that $tin2^{Acup B}$ makes $tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$?
discrete-mathematics elementary-set-theory
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
$endgroup$
add a comment |
$begingroup$
I need to prove that:
$2^{Acup B} =left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$
Note that $2^{A} = P(A)$ (The power set of A).
While I can prove the Right - Left direction fairly easily, It seems very hard for me to prove the Left - Right direction. Can I get some help with that?
How do I prove that $tin2^{Acup B}$ makes $tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$?
discrete-mathematics elementary-set-theory
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
$endgroup$
$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39
add a comment |
$begingroup$
I need to prove that:
$2^{Acup B} =left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$
Note that $2^{A} = P(A)$ (The power set of A).
While I can prove the Right - Left direction fairly easily, It seems very hard for me to prove the Left - Right direction. Can I get some help with that?
How do I prove that $tin2^{Acup B}$ makes $tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$?
discrete-mathematics elementary-set-theory
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
$endgroup$
I need to prove that:
$2^{Acup B} =left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$
Note that $2^{A} = P(A)$ (The power set of A).
While I can prove the Right - Left direction fairly easily, It seems very hard for me to prove the Left - Right direction. Can I get some help with that?
How do I prove that $tin2^{Acup B}$ makes $tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }$?
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
edited Mar 15 at 8:54
Marc van Leeuwen
88.5k5111229
88.5k5111229
asked Mar 15 at 8:32
trizztrizz
235
asked Mar 15 at 8:32
trizztrizz
235
asked Mar 15 at 8:32
trizztrizz
235
235
$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39
add a comment |
$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39
$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $tin2^{Acup B}$ then every element of $t$ is in $A$ or in $B$, so in either $tcap A$ or $tcap B$. Thus $t=Xcup Y$ where $X=tcap Ain2^A$ and $Y=tcap Bin 2^B$.
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
$endgroup$
add a comment |
$begingroup$
If $t in 2^{Acup B}$, then $ t subseteq A cup B.$
Put $t_1:= t cap A$ and $t_2:= t cap B$. Then $t_1 in 2^A, t_2 in 2^B $ and $t= t_1 cup t_2 $. Thus
$$tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }.$$
answered Mar 15 at 8:40
FredFred
48.8k11849
$endgroup$
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $tin2^{Acup B}$ then every element of $t$ is in $A$ or in $B$, so in either $tcap A$ or $tcap B$. Thus $t=Xcup Y$ where $X=tcap Ain2^A$ and $Y=tcap Bin 2^B$.
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
$endgroup$
add a comment |
$begingroup$
If $tin2^{Acup B}$ then every element of $t$ is in $A$ or in $B$, so in either $tcap A$ or $tcap B$. Thus $t=Xcup Y$ where $X=tcap Ain2^A$ and $Y=tcap Bin 2^B$.
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
$endgroup$
add a comment |
$begingroup$
If $tin2^{Acup B}$ then every element of $t$ is in $A$ or in $B$, so in either $tcap A$ or $tcap B$. Thus $t=Xcup Y$ where $X=tcap Ain2^A$ and $Y=tcap Bin 2^B$.
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
$endgroup$
If $tin2^{Acup B}$ then every element of $t$ is in $A$ or in $B$, so in either $tcap A$ or $tcap B$. Thus $t=Xcup Y$ where $X=tcap Ain2^A$ and $Y=tcap Bin 2^B$.
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
answered Mar 15 at 8:41
Especially LimeEspecially Lime
22.6k23059
22.6k23059
add a comment |
add a comment |
$begingroup$
If $t in 2^{Acup B}$, then $ t subseteq A cup B.$
Put $t_1:= t cap A$ and $t_2:= t cap B$. Then $t_1 in 2^A, t_2 in 2^B $ and $t= t_1 cup t_2 $. Thus
$$tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }.$$
answered Mar 15 at 8:40
FredFred
48.8k11849
$endgroup$
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
add a comment |
$begingroup$
If $t in 2^{Acup B}$, then $ t subseteq A cup B.$
Put $t_1:= t cap A$ and $t_2:= t cap B$. Then $t_1 in 2^A, t_2 in 2^B $ and $t= t_1 cup t_2 $. Thus
$$tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }.$$
answered Mar 15 at 8:40
FredFred
48.8k11849
$endgroup$
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
add a comment |
$begingroup$
If $t in 2^{Acup B}$, then $ t subseteq A cup B.$
Put $t_1:= t cap A$ and $t_2:= t cap B$. Then $t_1 in 2^A, t_2 in 2^B $ and $t= t_1 cup t_2 $. Thus
$$tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }.$$
answered Mar 15 at 8:40
FredFred
48.8k11849
$endgroup$
If $t in 2^{Acup B}$, then $ t subseteq A cup B.$
Put $t_1:= t cap A$ and $t_2:= t cap B$. Then $t_1 in 2^A, t_2 in 2^B $ and $t= t_1 cup t_2 $. Thus
$$tin left { Xcup Y mid Xin 2^{A}, Yin2^{B} right }.$$
answered Mar 15 at 8:40
FredFred
48.8k11849
answered Mar 15 at 8:40
FredFred
48.8k11849
answered Mar 15 at 8:40
FredFred
48.8k11849
answered Mar 15 at 8:40
FredFred
48.8k11849
48.8k11849
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
add a comment |
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
$begingroup$
Note that $t=t_1cup t_2$ again needs a proof showing both $subseteq$ and $supseteq$ (an easy one for sure).
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:47
add a comment |
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$begingroup$
I appears one is not supposing $A$ and $B$ disjoint here; the result is true without that hypothesis, but a bit more care in formulation is needed. I suppose "left-right" means the inclusion "$subseteq$". Also, it is not "power set", not "power group".
$endgroup$
– Marc van Leeuwen
Mar 15 at 8:42
$begingroup$
Possible duplicate of Prove $2^{(A cup B)}=left{ Xcup Y| Xin 2^A, Yin 2^Bright}$ for sets $A,B$.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 16:24
$begingroup$
Actually $2^A$ has a standard meaning - it's not literally equal to the power set. Here it actually makes a differennce..
$endgroup$
– David C. Ullrich
Mar 20 at 16:39