Mixing
Prove: $mathcal P(Acap B)=mathcal P(A)cap mathcal P(B)$
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Prove: $mathcal P(Acap B)=mathcal P(A)cap mathcal P(B)$
I tried it like this:
Let $Sin mathcal P(Acap B)implies Ssubseteq Acap Bimplies Ssubseteq A$ and $Ssubseteq Bimplies Sin mathcal P(A)$ and $Sin mathcal P(B)implies$
$$mathcal P(Acap B) = mathcal P(A)capmathcal P(B)$$
Can someone tell me if my proof is correct? and also I have a couple more questions on this topic:
What is the first approach, the first intuitive thought that you have when you need to prove something like this? For example what if the exercise was: $$mathcal P(A)cupmathcal P(B)subseteqmathcal P(Acup B)$$
What do you think about when you have this? Can someone explain me this in more detail and maybe using an example?
elementary-set-theory
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
$endgroup$
add a comment |
$begingroup$
Prove: $mathcal P(Acap B)=mathcal P(A)cap mathcal P(B)$
I tried it like this:
Let $Sin mathcal P(Acap B)implies Ssubseteq Acap Bimplies Ssubseteq A$ and $Ssubseteq Bimplies Sin mathcal P(A)$ and $Sin mathcal P(B)implies$
$$mathcal P(Acap B) = mathcal P(A)capmathcal P(B)$$
Can someone tell me if my proof is correct? and also I have a couple more questions on this topic:
What is the first approach, the first intuitive thought that you have when you need to prove something like this? For example what if the exercise was: $$mathcal P(A)cupmathcal P(B)subseteqmathcal P(Acup B)$$
What do you think about when you have this? Can someone explain me this in more detail and maybe using an example?
elementary-set-theory
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
$endgroup$
$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03
add a comment |
$begingroup$
Prove: $mathcal P(Acap B)=mathcal P(A)cap mathcal P(B)$
I tried it like this:
Let $Sin mathcal P(Acap B)implies Ssubseteq Acap Bimplies Ssubseteq A$ and $Ssubseteq Bimplies Sin mathcal P(A)$ and $Sin mathcal P(B)implies$
$$mathcal P(Acap B) = mathcal P(A)capmathcal P(B)$$
Can someone tell me if my proof is correct? and also I have a couple more questions on this topic:
What is the first approach, the first intuitive thought that you have when you need to prove something like this? For example what if the exercise was: $$mathcal P(A)cupmathcal P(B)subseteqmathcal P(Acup B)$$
What do you think about when you have this? Can someone explain me this in more detail and maybe using an example?
elementary-set-theory
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
$endgroup$
Prove: $mathcal P(Acap B)=mathcal P(A)cap mathcal P(B)$
I tried it like this:
Let $Sin mathcal P(Acap B)implies Ssubseteq Acap Bimplies Ssubseteq A$ and $Ssubseteq Bimplies Sin mathcal P(A)$ and $Sin mathcal P(B)implies$
$$mathcal P(Acap B) = mathcal P(A)capmathcal P(B)$$
Can someone tell me if my proof is correct? and also I have a couple more questions on this topic:
What is the first approach, the first intuitive thought that you have when you need to prove something like this? For example what if the exercise was: $$mathcal P(A)cupmathcal P(B)subseteqmathcal P(Acup B)$$
What do you think about when you have this? Can someone explain me this in more detail and maybe using an example?
elementary-set-theory
elementary-set-theory
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
edited Jan 17 at 12:29
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
asked Jan 17 at 7:26
C. CristiC. Cristi
1,634218
1,634218
$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03
add a comment |
$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03
$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a two-way proof of the first claim write $$xinmathcal{P}(Acap B)iff xsubseteq Acap Biff xsubseteq Aland xsubseteq Biff xinmathcal{P}(A)capmathcal{P}(B).$$(The approach you've already taken can flesh out the steps a little more, but the point is we can write the steps so each new statement is equivalent to the previous one, proving to directions at once.) For the second claim just replace every $cap,,land$ with $cup,,lor$. This works because of de Morgan's laws.
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
For a two-way proof of the first claim write $$xinmathcal{P}(Acap B)iff xsubseteq Acap Biff xsubseteq Aland xsubseteq Biff xinmathcal{P}(A)capmathcal{P}(B).$$(The approach you've already taken can flesh out the steps a little more, but the point is we can write the steps so each new statement is equivalent to the previous one, proving to directions at once.) For the second claim just replace every $cap,,land$ with $cup,,lor$. This works because of de Morgan's laws.
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
$begingroup$
For a two-way proof of the first claim write $$xinmathcal{P}(Acap B)iff xsubseteq Acap Biff xsubseteq Aland xsubseteq Biff xinmathcal{P}(A)capmathcal{P}(B).$$(The approach you've already taken can flesh out the steps a little more, but the point is we can write the steps so each new statement is equivalent to the previous one, proving to directions at once.) For the second claim just replace every $cap,,land$ with $cup,,lor$. This works because of de Morgan's laws.
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
$begingroup$
For a two-way proof of the first claim write $$xinmathcal{P}(Acap B)iff xsubseteq Acap Biff xsubseteq Aland xsubseteq Biff xinmathcal{P}(A)capmathcal{P}(B).$$(The approach you've already taken can flesh out the steps a little more, but the point is we can write the steps so each new statement is equivalent to the previous one, proving to directions at once.) For the second claim just replace every $cap,,land$ with $cup,,lor$. This works because of de Morgan's laws.
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
$endgroup$
For a two-way proof of the first claim write $$xinmathcal{P}(Acap B)iff xsubseteq Acap Biff xsubseteq Aland xsubseteq Biff xinmathcal{P}(A)capmathcal{P}(B).$$(The approach you've already taken can flesh out the steps a little more, but the point is we can write the steps so each new statement is equivalent to the previous one, proving to directions at once.) For the second claim just replace every $cap,,land$ with $cup,,lor$. This works because of de Morgan's laws.
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
answered Jan 17 at 7:43
J.G.J.G.
27.8k22843
27.8k22843
add a comment |
add a comment |
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$begingroup$
That looks like a proof of $P(Acap B)subseteq P(A)cap P(B)$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:27
$begingroup$
That's only half the proof. You also need to prove the implication the other way.
$endgroup$
– Michael Behrend
Jan 17 at 7:32
$begingroup$
Why was this question marked down? I don't see how it doesn't comply with the community guidelines? what have I missed?
$endgroup$
– DavidG
Jan 17 at 9:03