Mixing
When is A a subset of the power set p(A)?
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I can across a video (here's the pic: [Discrete Math 1] Subsets and Power Sets) and the author mentioned the following:
- Is A ⊆ p(A) for any A? No.
- Is A ∈ (A) for any A? Yes.
p(A) is the power set of A in this example. I agree with these two points mentioned.
However, I am confused with the exception he pointed out which is when p(∅). So we have p(A)={∅} and A=∅. It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
Is this because of ∅ being a subset of every set? If I go by this definition of a subset "The set A is a subset of the set B if and only if every element of A is also an element of B.", it does seem like this is one of the exceptions.
Is p({∅}) also an exception? A = {∅}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
Here's a picture from the video: [Discrete Math 1] Subsets and Power Sets
discrete-mathematics
asked Jan 15 at 14:21
supmethodssupmethods
72
$endgroup$
add a comment |
$begingroup$
I can across a video (here's the pic: [Discrete Math 1] Subsets and Power Sets) and the author mentioned the following:
- Is A ⊆ p(A) for any A? No.
- Is A ∈ (A) for any A? Yes.
p(A) is the power set of A in this example. I agree with these two points mentioned.
However, I am confused with the exception he pointed out which is when p(∅). So we have p(A)={∅} and A=∅. It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
Is this because of ∅ being a subset of every set? If I go by this definition of a subset "The set A is a subset of the set B if and only if every element of A is also an element of B.", it does seem like this is one of the exceptions.
Is p({∅}) also an exception? A = {∅}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
Here's a picture from the video: [Discrete Math 1] Subsets and Power Sets
discrete-mathematics
asked Jan 15 at 14:21
supmethodssupmethods
72
$endgroup$
add a comment |
$begingroup$
I can across a video (here's the pic: [Discrete Math 1] Subsets and Power Sets) and the author mentioned the following:
- Is A ⊆ p(A) for any A? No.
- Is A ∈ (A) for any A? Yes.
p(A) is the power set of A in this example. I agree with these two points mentioned.
However, I am confused with the exception he pointed out which is when p(∅). So we have p(A)={∅} and A=∅. It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
Is this because of ∅ being a subset of every set? If I go by this definition of a subset "The set A is a subset of the set B if and only if every element of A is also an element of B.", it does seem like this is one of the exceptions.
Is p({∅}) also an exception? A = {∅}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
Here's a picture from the video: [Discrete Math 1] Subsets and Power Sets
discrete-mathematics
asked Jan 15 at 14:21
supmethodssupmethods
72
$endgroup$
I can across a video (here's the pic: [Discrete Math 1] Subsets and Power Sets) and the author mentioned the following:
- Is A ⊆ p(A) for any A? No.
- Is A ∈ (A) for any A? Yes.
p(A) is the power set of A in this example. I agree with these two points mentioned.
However, I am confused with the exception he pointed out which is when p(∅). So we have p(A)={∅} and A=∅. It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
Is this because of ∅ being a subset of every set? If I go by this definition of a subset "The set A is a subset of the set B if and only if every element of A is also an element of B.", it does seem like this is one of the exceptions.
Is p({∅}) also an exception? A = {∅}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
Here's a picture from the video: [Discrete Math 1] Subsets and Power Sets
discrete-mathematics
discrete-mathematics
asked Jan 15 at 14:21
supmethodssupmethods
72
asked Jan 15 at 14:21
supmethodssupmethods
72
edited Jan 15 at 14:40
supmethods
asked Jan 15 at 14:21
supmethodssupmethods
72
asked Jan 15 at 14:21
supmethodssupmethods
72
asked Jan 15 at 14:21
supmethodssupmethods
72
72
add a comment |
add a comment |
2 Answers
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$begingroup$
If $A=emptyset$, then $Asubset P(A)$ is true, because $emptysetsubseteq{emptyset}$ is true. The latter is true because $emptysetsubseteq B$ is true for any set $B$.
Your statement, however, is confusing:
It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
The first part is not true. $A$ does not have an element. $emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $Asubseteq P(A)$, we see that $Asubseteq P(A)$ is also true.
Also, if $A={emptyset}$, then $P(A) = {emptyset, {emptyset}}$. In this case, again, $Asubseteq A$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.
Even more generally, taking $A=P(P(cdots P(emptyset)cdots))$, no matter how many powersets we take, we will always have $Asubseteq P(A)$.
EDIT:
I don't understand the video capture that you posted, but I will comment on this part:
Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
NO. You are not correct. If $A={{emptyset}}$, then $P(A)neq {emptyset,{emptyset}}$. Remember, $P(A)$ is the set of all subsets of $A$. ${emptyset}$ is not a subset of $A$, because it contains an element, $emptyset$, which is not an element of $A$.
The correct answer with $A={{emptyset}}$ is this:
For $A={{emptyset}}$, the statement $Asubseteq P(A)$ does not hold, however. This is because, for $A={{emptyset}}$, $P(A)={emptyset,{{emptyset}}}$. Now, there exists an element of $A$, in particular, ${emptyset}$, which is not an element of $P(A)$.
answered Jan 15 at 14:30
5xum5xum
90.9k394161
$endgroup$
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
add a comment |
$begingroup$
We have $Asubseteq p(A)$ if every element of $A$ is also a subset of $A$. That is if $xin yin A$ implies that also $xin A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $emptyset=:0$ and ${emptyset}=:1$.
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
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$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
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– 5xum
Jan 15 at 14:47
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
If $A=emptyset$, then $Asubset P(A)$ is true, because $emptysetsubseteq{emptyset}$ is true. The latter is true because $emptysetsubseteq B$ is true for any set $B$.
Your statement, however, is confusing:
It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
The first part is not true. $A$ does not have an element. $emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $Asubseteq P(A)$, we see that $Asubseteq P(A)$ is also true.
Also, if $A={emptyset}$, then $P(A) = {emptyset, {emptyset}}$. In this case, again, $Asubseteq A$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.
Even more generally, taking $A=P(P(cdots P(emptyset)cdots))$, no matter how many powersets we take, we will always have $Asubseteq P(A)$.
EDIT:
I don't understand the video capture that you posted, but I will comment on this part:
Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
NO. You are not correct. If $A={{emptyset}}$, then $P(A)neq {emptyset,{emptyset}}$. Remember, $P(A)$ is the set of all subsets of $A$. ${emptyset}$ is not a subset of $A$, because it contains an element, $emptyset$, which is not an element of $A$.
The correct answer with $A={{emptyset}}$ is this:
For $A={{emptyset}}$, the statement $Asubseteq P(A)$ does not hold, however. This is because, for $A={{emptyset}}$, $P(A)={emptyset,{{emptyset}}}$. Now, there exists an element of $A$, in particular, ${emptyset}$, which is not an element of $P(A)$.
answered Jan 15 at 14:30
5xum5xum
90.9k394161
$endgroup$
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
add a comment |
$begingroup$
If $A=emptyset$, then $Asubset P(A)$ is true, because $emptysetsubseteq{emptyset}$ is true. The latter is true because $emptysetsubseteq B$ is true for any set $B$.
Your statement, however, is confusing:
It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
The first part is not true. $A$ does not have an element. $emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $Asubseteq P(A)$, we see that $Asubseteq P(A)$ is also true.
Also, if $A={emptyset}$, then $P(A) = {emptyset, {emptyset}}$. In this case, again, $Asubseteq A$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.
Even more generally, taking $A=P(P(cdots P(emptyset)cdots))$, no matter how many powersets we take, we will always have $Asubseteq P(A)$.
EDIT:
I don't understand the video capture that you posted, but I will comment on this part:
Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
NO. You are not correct. If $A={{emptyset}}$, then $P(A)neq {emptyset,{emptyset}}$. Remember, $P(A)$ is the set of all subsets of $A$. ${emptyset}$ is not a subset of $A$, because it contains an element, $emptyset$, which is not an element of $A$.
The correct answer with $A={{emptyset}}$ is this:
For $A={{emptyset}}$, the statement $Asubseteq P(A)$ does not hold, however. This is because, for $A={{emptyset}}$, $P(A)={emptyset,{{emptyset}}}$. Now, there exists an element of $A$, in particular, ${emptyset}$, which is not an element of $P(A)$.
answered Jan 15 at 14:30
5xum5xum
90.9k394161
$endgroup$
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
add a comment |
$begingroup$
If $A=emptyset$, then $Asubset P(A)$ is true, because $emptysetsubseteq{emptyset}$ is true. The latter is true because $emptysetsubseteq B$ is true for any set $B$.
Your statement, however, is confusing:
It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
The first part is not true. $A$ does not have an element. $emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $Asubseteq P(A)$, we see that $Asubseteq P(A)$ is also true.
Also, if $A={emptyset}$, then $P(A) = {emptyset, {emptyset}}$. In this case, again, $Asubseteq A$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.
Even more generally, taking $A=P(P(cdots P(emptyset)cdots))$, no matter how many powersets we take, we will always have $Asubseteq P(A)$.
EDIT:
I don't understand the video capture that you posted, but I will comment on this part:
Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
NO. You are not correct. If $A={{emptyset}}$, then $P(A)neq {emptyset,{emptyset}}$. Remember, $P(A)$ is the set of all subsets of $A$. ${emptyset}$ is not a subset of $A$, because it contains an element, $emptyset$, which is not an element of $A$.
The correct answer with $A={{emptyset}}$ is this:
For $A={{emptyset}}$, the statement $Asubseteq P(A)$ does not hold, however. This is because, for $A={{emptyset}}$, $P(A)={emptyset,{{emptyset}}}$. Now, there exists an element of $A$, in particular, ${emptyset}$, which is not an element of $P(A)$.
answered Jan 15 at 14:30
5xum5xum
90.9k394161
$endgroup$
If $A=emptyset$, then $Asubset P(A)$ is true, because $emptysetsubseteq{emptyset}$ is true. The latter is true because $emptysetsubseteq B$ is true for any set $B$.
Your statement, however, is confusing:
It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).
The first part is not true. $A$ does not have an element. $emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $Asubseteq P(A)$, we see that $Asubseteq P(A)$ is also true.
Also, if $A={emptyset}$, then $P(A) = {emptyset, {emptyset}}$. In this case, again, $Asubseteq A$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.
Even more generally, taking $A=P(P(cdots P(emptyset)cdots))$, no matter how many powersets we take, we will always have $Asubseteq P(A)$.
EDIT:
I don't understand the video capture that you posted, but I will comment on this part:
Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
NO. You are not correct. If $A={{emptyset}}$, then $P(A)neq {emptyset,{emptyset}}$. Remember, $P(A)$ is the set of all subsets of $A$. ${emptyset}$ is not a subset of $A$, because it contains an element, $emptyset$, which is not an element of $A$.
The correct answer with $A={{emptyset}}$ is this:
For $A={{emptyset}}$, the statement $Asubseteq P(A)$ does not hold, however. This is because, for $A={{emptyset}}$, $P(A)={emptyset,{{emptyset}}}$. Now, there exists an element of $A$, in particular, ${emptyset}$, which is not an element of $P(A)$.
answered Jan 15 at 14:30
5xum5xum
90.9k394161
edited Jan 15 at 14:42
answered Jan 15 at 14:30
5xum5xum
90.9k394161
answered Jan 15 at 14:30
5xum5xum
90.9k394161
answered Jan 15 at 14:30
5xum5xum
90.9k394161
90.9k394161
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
add a comment |
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
I made some corrections and posted an image of the video snippet. I believe A = {∅} in the first example and A = {{∅}}. Please correct me if I am correct. Thanks.
$endgroup$
– supmethods
Jan 15 at 14:36
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
@supmethods I edited my answer.
$endgroup$
– 5xum
Jan 15 at 14:42
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
P(P(⋯P(∅)⋯)) = {∅} right?
$endgroup$
– supmethods
Jan 15 at 15:02
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
$begingroup$
Ok, I think I got it. A = P(P(∅)) = {∅, {∅}}, so P(A) = {∅, {∅}, {{∅}}, {∅,{∅}}} and A⊆P(A).
$endgroup$
– supmethods
Jan 15 at 15:23
add a comment |
$begingroup$
We have $Asubseteq p(A)$ if every element of $A$ is also a subset of $A$. That is if $xin yin A$ implies that also $xin A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $emptyset=:0$ and ${emptyset}=:1$.
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
add a comment |
$begingroup$
We have $Asubseteq p(A)$ if every element of $A$ is also a subset of $A$. That is if $xin yin A$ implies that also $xin A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $emptyset=:0$ and ${emptyset}=:1$.
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
add a comment |
$begingroup$
We have $Asubseteq p(A)$ if every element of $A$ is also a subset of $A$. That is if $xin yin A$ implies that also $xin A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $emptyset=:0$ and ${emptyset}=:1$.
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
We have $Asubseteq p(A)$ if every element of $A$ is also a subset of $A$. That is if $xin yin A$ implies that also $xin A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $emptyset=:0$ and ${emptyset}=:1$.
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:30
Hagen von EitzenHagen von Eitzen
1643
1643
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
add a comment |
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
$begingroup$
math.stackexchange.com/users/573610/hagen-von-eitzen math.stackexchange.com/users/39174/hagen-von-eitzen ?
$endgroup$
– 5xum
Jan 15 at 14:47
add a comment |
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