Mixing
Operation on Sets - Listing the elements
$begingroup$
Encountered a question on set operations and am kind of lost...
$(D bigcup {A})oplus A$
$A: {a,b,c,d}, D: {b,d}$
Does ${A}$ mean ${{a,b,c,d}}$, making it a set of a set? Would set D union set of set A (?) be the same as set D union set A, or are they different?
I would understand how to list the elements if it were $(D bigcup A)oplus A$, but the ${A}$ is what is confusing me.
discrete-mathematics elementary-set-theory
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
$endgroup$
add a comment |
$begingroup$
Encountered a question on set operations and am kind of lost...
$(D bigcup {A})oplus A$
$A: {a,b,c,d}, D: {b,d}$
Does ${A}$ mean ${{a,b,c,d}}$, making it a set of a set? Would set D union set of set A (?) be the same as set D union set A, or are they different?
I would understand how to list the elements if it were $(D bigcup A)oplus A$, but the ${A}$ is what is confusing me.
discrete-mathematics elementary-set-theory
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
$endgroup$
$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44
add a comment |
$begingroup$
Encountered a question on set operations and am kind of lost...
$(D bigcup {A})oplus A$
$A: {a,b,c,d}, D: {b,d}$
Does ${A}$ mean ${{a,b,c,d}}$, making it a set of a set? Would set D union set of set A (?) be the same as set D union set A, or are they different?
I would understand how to list the elements if it were $(D bigcup A)oplus A$, but the ${A}$ is what is confusing me.
discrete-mathematics elementary-set-theory
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
$endgroup$
Encountered a question on set operations and am kind of lost...
$(D bigcup {A})oplus A$
$A: {a,b,c,d}, D: {b,d}$
Does ${A}$ mean ${{a,b,c,d}}$, making it a set of a set? Would set D union set of set A (?) be the same as set D union set A, or are they different?
I would understand how to list the elements if it were $(D bigcup A)oplus A$, but the ${A}$ is what is confusing me.
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
edited Jan 28 at 11:34
J. W. Tanner
3,9221320
3,9221320
asked Jan 28 at 7:39
BandoleroBandolero
243
asked Jan 28 at 7:39
BandoleroBandolero
243
asked Jan 28 at 7:39
BandoleroBandolero
243
243
$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44
add a comment |
$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44
$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Good question!
However, you should write
$$A = {a,b,c,d},D = {b,d},$$ because after all, $D$ is being defined to equal the set ${b,d}$, so why use a symbol other than the equality symbol to denote equality in this case?
You're correct that ${A}$ means ${{a,b,c,d}}$. Hence $$D cup {A} = {{a,b,c,d},b,d},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Good question!
However, you should write
$$A = {a,b,c,d},D = {b,d},$$ because after all, $D$ is being defined to equal the set ${b,d}$, so why use a symbol other than the equality symbol to denote equality in this case?
You're correct that ${A}$ means ${{a,b,c,d}}$. Hence $$D cup {A} = {{a,b,c,d},b,d},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
add a comment |
$begingroup$
Good question!
However, you should write
$$A = {a,b,c,d},D = {b,d},$$ because after all, $D$ is being defined to equal the set ${b,d}$, so why use a symbol other than the equality symbol to denote equality in this case?
You're correct that ${A}$ means ${{a,b,c,d}}$. Hence $$D cup {A} = {{a,b,c,d},b,d},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
add a comment |
$begingroup$
Good question!
However, you should write
$$A = {a,b,c,d},D = {b,d},$$ because after all, $D$ is being defined to equal the set ${b,d}$, so why use a symbol other than the equality symbol to denote equality in this case?
You're correct that ${A}$ means ${{a,b,c,d}}$. Hence $$D cup {A} = {{a,b,c,d},b,d},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
$endgroup$
Good question!
However, you should write
$$A = {a,b,c,d},D = {b,d},$$ because after all, $D$ is being defined to equal the set ${b,d}$, so why use a symbol other than the equality symbol to denote equality in this case?
You're correct that ${A}$ means ${{a,b,c,d}}$. Hence $$D cup {A} = {{a,b,c,d},b,d},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
answered Jan 28 at 7:59
goblingoblin
37.1k1159193
37.1k1159193
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
add a comment |
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
Ok, so from what I understand, it should be ${{a,b,c,d}b,d}oplus{a,b,c,d}$. still not sure about the answer tho, but I'm guessing it would be between ${{a,b,c,d}a,c}$ or ${a,c}$
$endgroup$
– Bandolero
Jan 28 at 8:13
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
@Bandolero, to find $P oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets.
$endgroup$
– goblin
Jan 28 at 8:20
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
I got that part, but my question is would say ${P}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep ${P}$ untouched?
$endgroup$
– Bandolero
Jan 28 at 8:25
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
$begingroup$
@Bandolero, I'm not sure I understand, but ${P}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and ${P}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory.
$endgroup$
– goblin
Jan 30 at 11:58
add a comment |
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$begingroup$
What does $oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union?
$endgroup$
– goblin
Jan 28 at 7:44
$begingroup$
Symmetric difference, so if A⊕B, elements that are in A or B but not both
$endgroup$
– Bandolero
Jan 28 at 7:55
$begingroup$
@Bandolero the usual notation for symmetric difference is $AΔB$
$endgroup$
– Holo
Jan 28 at 11:44