Mixing
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Does the power set of a set contain a copy of the elements in the set?
Clash Royale CLAN TAG#URR8PPP
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If A is a set and P(A) its power set, then P(A) contains all the elements in A. P(A) also contains the subsets of A, which A doesn't necessarily contain.
So, does P(A) contain a copy of all the elements in A, and not the same elements?
Hence, if A = x is a singleton, B = x ⊆ P(A), then A and B are not the same element, despite A = B. Can this be clarified to me?
elementary-set-theory inclusion-exclusion
asked Jul 28 at 3:03
davide
1155
add a comment |Â
up vote
-2
down vote
favorite
If A is a set and P(A) its power set, then P(A) contains all the elements in A. P(A) also contains the subsets of A, which A doesn't necessarily contain.
So, does P(A) contain a copy of all the elements in A, and not the same elements?
Hence, if A = x is a singleton, B = x ⊆ P(A), then A and B are not the same element, despite A = B. Can this be clarified to me?
elementary-set-theory inclusion-exclusion
asked Jul 28 at 3:03
davide
1155
2
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If A is a set and P(A) its power set, then P(A) contains all the elements in A. P(A) also contains the subsets of A, which A doesn't necessarily contain.
So, does P(A) contain a copy of all the elements in A, and not the same elements?
Hence, if A = x is a singleton, B = x ⊆ P(A), then A and B are not the same element, despite A = B. Can this be clarified to me?
elementary-set-theory inclusion-exclusion
asked Jul 28 at 3:03
davide
1155
If A is a set and P(A) its power set, then P(A) contains all the elements in A. P(A) also contains the subsets of A, which A doesn't necessarily contain.
So, does P(A) contain a copy of all the elements in A, and not the same elements?
Hence, if A = x is a singleton, B = x ⊆ P(A), then A and B are not the same element, despite A = B. Can this be clarified to me?
elementary-set-theory inclusion-exclusion
asked Jul 28 at 3:03
davide
1155
edited Jul 28 at 3:11
asked Jul 28 at 3:03
davide
1155
asked Jul 28 at 3:03
davide
1155
asked Jul 28 at 3:03
davide
1155
1155
2
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29
add a comment |Â
2
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29
2
2
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $Asubseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=$apple$$ is $P(A)=emptyset, $apple$$.
However, apple isn't an element of $P(A)$: $emptysetnot=$ apple and $$apple$not=$ apple.
I think the key point is this last bit: in general, we won't have $x=x$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $emptyset$ versus $emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $ain A$, the set $a$ is an element of $P(A)$, and the map $Arightarrow P(A):amapsto a$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $a$. But this is not the same thing as the object $a$.
answered Jul 28 at 3:12
Noah Schweber
110k9139260
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to asa, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.
– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $Asubseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=$apple$$ is $P(A)=emptyset, $apple$$.
However, apple isn't an element of $P(A)$: $emptysetnot=$ apple and $$apple$not=$ apple.
I think the key point is this last bit: in general, we won't have $x=x$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $emptyset$ versus $emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $ain A$, the set $a$ is an element of $P(A)$, and the map $Arightarrow P(A):amapsto a$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $a$. But this is not the same thing as the object $a$.
answered Jul 28 at 3:12
Noah Schweber
110k9139260
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to asa, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.
– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
add a comment |Â
up vote
3
down vote
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $Asubseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=$apple$$ is $P(A)=emptyset, $apple$$.
However, apple isn't an element of $P(A)$: $emptysetnot=$ apple and $$apple$not=$ apple.
I think the key point is this last bit: in general, we won't have $x=x$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $emptyset$ versus $emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $ain A$, the set $a$ is an element of $P(A)$, and the map $Arightarrow P(A):amapsto a$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $a$. But this is not the same thing as the object $a$.
answered Jul 28 at 3:12
Noah Schweber
110k9139260
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to asa, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.
– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $Asubseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=$apple$$ is $P(A)=emptyset, $apple$$.
However, apple isn't an element of $P(A)$: $emptysetnot=$ apple and $$apple$not=$ apple.
I think the key point is this last bit: in general, we won't have $x=x$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $emptyset$ versus $emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $ain A$, the set $a$ is an element of $P(A)$, and the map $Arightarrow P(A):amapsto a$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $a$. But this is not the same thing as the object $a$.
answered Jul 28 at 3:12
Noah Schweber
110k9139260
I'm assuming here that "$P(A)$" means "the powerset of $A$."
Your claim that in general $Asubseteq P(A)$ is false. For simplicity, let's work in naive set theory (although everything I say is also true in, say, ZFC). Then:
The powerset of the set $A=$apple$$ is $P(A)=emptyset, $apple$$.
However, apple isn't an element of $P(A)$: $emptysetnot=$ apple and $$apple$not=$ apple.
I think the key point is this last bit: in general, we won't have $x=x$. (Indeed, in ZFC we never will have this, by the axiom of regularity!) The ur-example is $emptyset$ versus $emptyset$ (one of these isn't empty!).
What is true is that there is a natural injection of $A$ into $P(A)$: for $ain A$, the set $a$ is an element of $P(A)$, and the map $Arightarrow P(A):amapsto a$ (the "singleton map") is injective. That is, for each element $a$ of $A$, the powerset $P(A)$ contains something corresponding to the object $a$, namely $a$. But this is not the same thing as the object $a$.
answered Jul 28 at 3:12
Noah Schweber
110k9139260
answered Jul 28 at 3:12
Noah Schweber
110k9139260
answered Jul 28 at 3:12
Noah Schweber
110k9139260
answered Jul 28 at 3:12
Noah Schweber
110k9139260
110k9139260
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to asa, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.
– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
add a comment |Â
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to asa, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.
– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to as
a, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.– davide
Jul 28 at 3:41
Thank you. Since P(A) is disjoint from A, is the element a∈A the same element a∈a∈P(A)? I don't understand if both elements referred to as
a, contained in two disjoint sets, can be the same element, of if they are copies of each other. Up until now my books left ambiguity on this concept.– davide
Jul 28 at 3:41
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
@davide Note that I didn't say that $P(A)$ is always disjoint from $A$; e.g. if we take $A=emptyset$, then $P(A)=emptyset,emptyset$ and $emptyset$ is an element of both $A$ and $P(A)$. However, let's for simplicity look at the case $A=a$, where $anot=emptyset$ and $anot=a$ (these stipulations ensure that $A$ and $P(A)$ are disjoint; the second is guaranteed if we assume the axiom of regularity, but some set theories don't have that). The objects $a$ and $a$ are distinct, and we have $ain a$ and $ain P(A)$ but $anotin P(A)$. (cont'd)
– Noah Schweber
Jul 28 at 3:59
1
1
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
I'm not sure what "copies of" means. We have $A=a$ and $P(A)=emptyset,a$; every $a$ is the same as every other $a$, but $a$ and $a$ are different things. Does this help?
– Noah Schweber
Jul 28 at 4:00
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
Thank you again. I was probably influenced by the fact that in computer science every element belongs to a memory location, and if two distinct locations contain the same element, then they are equal copies of each other.
– davide
Jul 28 at 4:07
add a comment |Â
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2
What is a "part of a set"?
– parsiad
Jul 28 at 3:07
Let "parts of a set" be "power set". Sorry, bad translation, fixing now.
– davide
Jul 28 at 3:09
Sorry for confusion, I was clarified that if A = a , then a∉A. My books don't make this exclusion explicit.
– davide
Jul 28 at 4:29