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The existence of onto function of a set to its power sets.
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Cantor’s Theorem: Given any set $A$, there does not exist a function $f : A → P(A)$ that is onto.
The book that I am reading says:
Assume, for contradiction, that, $f$ : $A → P (A)$ is onto. Where $f(a)$ is a particular subset of $A$, for each element of A. To arrive at a contradiction, we will produce a subset B ⊆ A that is not equal to f(a) for any $a ∈ A$.
$B = $ a ∈ A : a does not belong to f ( a ) .
Now assume that there is some $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$, then $a'$ does NOT belong to $B$, hence contradiction. Similarly for the case when $a'$ doesn't belong to $B$.
The question:
Having $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$ why does it imply that $a'$ does NOT belong to $B$? After all it is the definition of $B$, to consist the elements that does NOT belong to $f(a)$ for any $a$ from $A$. In this case $a'$.
P.S. I saw similar questions about Cantor's Theorem but didn't find the answer to my question.
elementary-set-theory
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
add a comment |Â
up vote
-2
down vote
favorite
Cantor’s Theorem: Given any set $A$, there does not exist a function $f : A → P(A)$ that is onto.
The book that I am reading says:
Assume, for contradiction, that, $f$ : $A → P (A)$ is onto. Where $f(a)$ is a particular subset of $A$, for each element of A. To arrive at a contradiction, we will produce a subset B ⊆ A that is not equal to f(a) for any $a ∈ A$.
$B = $ a ∈ A : a does not belong to f ( a ) .
Now assume that there is some $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$, then $a'$ does NOT belong to $B$, hence contradiction. Similarly for the case when $a'$ doesn't belong to $B$.
The question:
Having $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$ why does it imply that $a'$ does NOT belong to $B$? After all it is the definition of $B$, to consist the elements that does NOT belong to $f(a)$ for any $a$ from $A$. In this case $a'$.
P.S. I saw similar questions about Cantor's Theorem but didn't find the answer to my question.
elementary-set-theory
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Cantor’s Theorem: Given any set $A$, there does not exist a function $f : A → P(A)$ that is onto.
The book that I am reading says:
Assume, for contradiction, that, $f$ : $A → P (A)$ is onto. Where $f(a)$ is a particular subset of $A$, for each element of A. To arrive at a contradiction, we will produce a subset B ⊆ A that is not equal to f(a) for any $a ∈ A$.
$B = $ a ∈ A : a does not belong to f ( a ) .
Now assume that there is some $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$, then $a'$ does NOT belong to $B$, hence contradiction. Similarly for the case when $a'$ doesn't belong to $B$.
The question:
Having $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$ why does it imply that $a'$ does NOT belong to $B$? After all it is the definition of $B$, to consist the elements that does NOT belong to $f(a)$ for any $a$ from $A$. In this case $a'$.
P.S. I saw similar questions about Cantor's Theorem but didn't find the answer to my question.
elementary-set-theory
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
Cantor’s Theorem: Given any set $A$, there does not exist a function $f : A → P(A)$ that is onto.
The book that I am reading says:
Assume, for contradiction, that, $f$ : $A → P (A)$ is onto. Where $f(a)$ is a particular subset of $A$, for each element of A. To arrive at a contradiction, we will produce a subset B ⊆ A that is not equal to f(a) for any $a ∈ A$.
$B = $ a ∈ A : a does not belong to f ( a ) .
Now assume that there is some $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$, then $a'$ does NOT belong to $B$, hence contradiction. Similarly for the case when $a'$ doesn't belong to $B$.
The question:
Having $a'$ ∈ $B$, then $a'$ does NOT belong to $f(a')$ why does it imply that $a'$ does NOT belong to $B$? After all it is the definition of $B$, to consist the elements that does NOT belong to $f(a)$ for any $a$ from $A$. In this case $a'$.
P.S. I saw similar questions about Cantor's Theorem but didn't find the answer to my question.
elementary-set-theory
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
asked Jul 26 at 17:38
Sargis Iskandaryan
47112
47112
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The contradiction arises if you assume that $B=f(a)$ for some $ain A$.
Then: $$ain f(a)iff anotin f(a)$$ The conclusion is that no such $a$ exists.
answered Jul 26 at 17:50
drhab
86.1k541118
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The contradiction arises if you assume that $B=f(a)$ for some $ain A$.
Then: $$ain f(a)iff anotin f(a)$$ The conclusion is that no such $a$ exists.
answered Jul 26 at 17:50
drhab
86.1k541118
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
add a comment |Â
up vote
1
down vote
accepted
The contradiction arises if you assume that $B=f(a)$ for some $ain A$.
Then: $$ain f(a)iff anotin f(a)$$ The conclusion is that no such $a$ exists.
answered Jul 26 at 17:50
drhab
86.1k541118
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The contradiction arises if you assume that $B=f(a)$ for some $ain A$.
Then: $$ain f(a)iff anotin f(a)$$ The conclusion is that no such $a$ exists.
answered Jul 26 at 17:50
drhab
86.1k541118
The contradiction arises if you assume that $B=f(a)$ for some $ain A$.
Then: $$ain f(a)iff anotin f(a)$$ The conclusion is that no such $a$ exists.
answered Jul 26 at 17:50
drhab
86.1k541118
answered Jul 26 at 17:50
drhab
86.1k541118
answered Jul 26 at 17:50
drhab
86.1k541118
answered Jul 26 at 17:50
drhab
86.1k541118
86.1k541118
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
add a comment |Â
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
But the assumption is that B = a ∈ A : a does not belong to f ( a ) and by this assumption if a' does not belong to f(a'), then it belongs to B. Can you, please, explain?
– Sargis Iskandaryan
Jul 26 at 17:54
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
And if $B=f(a)$ and $a'∈B$ doesn't it mean that $a'∈f(a')$? However, by assumption it does not belong to $f(a')$. Is this the contradiction? So, there is an additional assumption by setting $B=f(a)$?
– Sargis Iskandaryan
Jul 26 at 17:59
1
1
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
The main assumption must be that some $ain A$ exists such that $f(a)=xin Amid xnotin f(x)$. That leads to $ain f(a)iff anotin f(a)$ which is absurd. Do you see that? I rather do not judge about the book you are reading. Maybe you misinterpreted.
– drhab
Jul 26 at 18:00
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
Ok, in words $f(a)$ maps any element $x$ from $A$ to its subset(power set) such that this element does not belong to its subset $f(x)$. Is my interpretation right?
– Sargis Iskandaryan
Jul 26 at 18:25
1
1
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
In words: assume that a surjective function $f:Atowp(A)$ exists. Now define $B:=xin Amid xnotin f(x)$. Since $f$ is surjective there must be an $ain A$ with $f(a)=B$. That however leads to $ain f(a)iff anotin f(a)$ which is absurd. So the assumption that a surjective function $f:Atowp(A)$ exists must be false.
– drhab
Jul 26 at 19:45
add a comment |Â
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