Proof using set theory
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Consider the function 𝑓: 𝐴 → 𝐵 and the statement
𝑃(𝐶,𝐷):∀𝐶,𝐷 ∈ 𝒫(𝐵), 𝑓^−1 = (𝐶 ∩ 𝐷^𝑐)= 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c
where 𝒫(𝐵) is the power set of set.
𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷) ≡ ∀𝐶, 𝐷 ∈ 𝒫(𝐵), 𝑓^-1 (C-D) = f^-1(c) - [f^-1(D)]^c
You must use the set-element method of proof to prove that 𝑃(𝐶,𝐷) is true. In your proof you must consider all possible subset relationships between 𝐶 and 𝐷, including cases
of empty sets.
I cannot figure out what sets I start out with and why it says 𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷). Do I use C and D^c or A and B (if it is the latter I don't quite understand how the proof would be done)? Also how would I define all the cases (I generally know what they are but would I write for example case 1: x is an element of C intersect D^c, C and D^c are not the empty set) ?
For the first case, I am considering where x is an element of the intersection of C and D^c. Can I assume that if x is in this intersection then x is also in f^-1(C ∩ D^c), and then that by definition of f^-1, x is also in 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c ? From there the set difference law would get us to f^-1(c) - [f^-1(D)]^c, but I don't know if this is a correct proof or if I'm actually answering what the question is asking.
Thanks
elementary-set-theory
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
add a comment |
up vote
0
down vote
favorite
Consider the function 𝑓: 𝐴 → 𝐵 and the statement
𝑃(𝐶,𝐷):∀𝐶,𝐷 ∈ 𝒫(𝐵), 𝑓^−1 = (𝐶 ∩ 𝐷^𝑐)= 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c
where 𝒫(𝐵) is the power set of set.
𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷) ≡ ∀𝐶, 𝐷 ∈ 𝒫(𝐵), 𝑓^-1 (C-D) = f^-1(c) - [f^-1(D)]^c
You must use the set-element method of proof to prove that 𝑃(𝐶,𝐷) is true. In your proof you must consider all possible subset relationships between 𝐶 and 𝐷, including cases
of empty sets.
I cannot figure out what sets I start out with and why it says 𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷). Do I use C and D^c or A and B (if it is the latter I don't quite understand how the proof would be done)? Also how would I define all the cases (I generally know what they are but would I write for example case 1: x is an element of C intersect D^c, C and D^c are not the empty set) ?
For the first case, I am considering where x is an element of the intersection of C and D^c. Can I assume that if x is in this intersection then x is also in f^-1(C ∩ D^c), and then that by definition of f^-1, x is also in 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c ? From there the set difference law would get us to f^-1(c) - [f^-1(D)]^c, but I don't know if this is a correct proof or if I'm actually answering what the question is asking.
Thanks
elementary-set-theory
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the function 𝑓: 𝐴 → 𝐵 and the statement
𝑃(𝐶,𝐷):∀𝐶,𝐷 ∈ 𝒫(𝐵), 𝑓^−1 = (𝐶 ∩ 𝐷^𝑐)= 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c
where 𝒫(𝐵) is the power set of set.
𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷) ≡ ∀𝐶, 𝐷 ∈ 𝒫(𝐵), 𝑓^-1 (C-D) = f^-1(c) - [f^-1(D)]^c
You must use the set-element method of proof to prove that 𝑃(𝐶,𝐷) is true. In your proof you must consider all possible subset relationships between 𝐶 and 𝐷, including cases
of empty sets.
I cannot figure out what sets I start out with and why it says 𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷). Do I use C and D^c or A and B (if it is the latter I don't quite understand how the proof would be done)? Also how would I define all the cases (I generally know what they are but would I write for example case 1: x is an element of C intersect D^c, C and D^c are not the empty set) ?
For the first case, I am considering where x is an element of the intersection of C and D^c. Can I assume that if x is in this intersection then x is also in f^-1(C ∩ D^c), and then that by definition of f^-1, x is also in 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c ? From there the set difference law would get us to f^-1(c) - [f^-1(D)]^c, but I don't know if this is a correct proof or if I'm actually answering what the question is asking.
Thanks
elementary-set-theory
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
Consider the function 𝑓: 𝐴 → 𝐵 and the statement
𝑃(𝐶,𝐷):∀𝐶,𝐷 ∈ 𝒫(𝐵), 𝑓^−1 = (𝐶 ∩ 𝐷^𝑐)= 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c
where 𝒫(𝐵) is the power set of set.
𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷) ≡ ∀𝐶, 𝐷 ∈ 𝒫(𝐵), 𝑓^-1 (C-D) = f^-1(c) - [f^-1(D)]^c
You must use the set-element method of proof to prove that 𝑃(𝐶,𝐷) is true. In your proof you must consider all possible subset relationships between 𝐶 and 𝐷, including cases
of empty sets.
I cannot figure out what sets I start out with and why it says 𝑃(𝐶,𝐷) ∧ 𝑃(𝐶, 𝐷). Do I use C and D^c or A and B (if it is the latter I don't quite understand how the proof would be done)? Also how would I define all the cases (I generally know what they are but would I write for example case 1: x is an element of C intersect D^c, C and D^c are not the empty set) ?
For the first case, I am considering where x is an element of the intersection of C and D^c. Can I assume that if x is in this intersection then x is also in f^-1(C ∩ D^c), and then that by definition of f^-1, x is also in 𝑓^−1(𝐶) ∩ [𝑓^−1(𝐷)]^c ? From there the set difference law would get us to f^-1(c) - [f^-1(D)]^c, but I don't know if this is a correct proof or if I'm actually answering what the question is asking.
Thanks
elementary-set-theory
elementary-set-theory
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
edited Nov 15 at 0:54
Andrés E. Caicedo
64.2k8157243
64.2k8157243
asked Nov 14 at 22:49
Deb Martin
61
asked Nov 14 at 22:49
Deb Martin
61
asked Nov 14 at 22:49
Deb Martin
61
61
Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18
add a comment |
Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18
Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18
Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18
add a comment |
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Most of the variables show as empty rectangles making your question unreadable.
– William Elliot
Nov 15 at 9:18