Mixing
Existence of set $B={xin Amid xnotin f(x)}$ in Cantor's theorem.
-3
$begingroup$
For the proof in wiki:
https://en.m.wikipedia.org/wiki/Cantor%27s_theorem
$B={xin Amid xnotin f(x)}$
Example, where $B$ does NOT exist:
$A={a,b}$
$P(A)={{a},{b},{a,b},emptyset}$
Possible correspondence:
$aleftrightarrow {a}$,
$bleftrightarrow {b}$
So, $B$ does not always exist. So, the proof does not always work.
elementary-set-theory
edited Jan 20 at 10:39
Henno Brandsma
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
$endgroup$
|
show 4 more comments
-3
$begingroup$
For the proof in wiki:
https://en.m.wikipedia.org/wiki/Cantor%27s_theorem
$B={xin Amid xnotin f(x)}$
Example, where $B$ does NOT exist:
$A={a,b}$
$P(A)={{a},{b},{a,b},emptyset}$
Possible correspondence:
$aleftrightarrow {a}$,
$bleftrightarrow {b}$
So, $B$ does not always exist. So, the proof does not always work.
elementary-set-theory
edited Jan 20 at 10:39
Henno Brandsma
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
$endgroup$
$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
1
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
1
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21
|
show 4 more comments
-3
-3
-3
$begingroup$
For the proof in wiki:
https://en.m.wikipedia.org/wiki/Cantor%27s_theorem
$B={xin Amid xnotin f(x)}$
Example, where $B$ does NOT exist:
$A={a,b}$
$P(A)={{a},{b},{a,b},emptyset}$
Possible correspondence:
$aleftrightarrow {a}$,
$bleftrightarrow {b}$
So, $B$ does not always exist. So, the proof does not always work.
elementary-set-theory
edited Jan 20 at 10:39
Henno Brandsma
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
$endgroup$
For the proof in wiki:
https://en.m.wikipedia.org/wiki/Cantor%27s_theorem
$B={xin Amid xnotin f(x)}$
Example, where $B$ does NOT exist:
$A={a,b}$
$P(A)={{a},{b},{a,b},emptyset}$
Possible correspondence:
$aleftrightarrow {a}$,
$bleftrightarrow {b}$
So, $B$ does not always exist. So, the proof does not always work.
elementary-set-theory
elementary-set-theory
edited Jan 20 at 10:39
Henno Brandsma
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
edited Jan 20 at 10:39
Henno Brandsma
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
edited Jan 20 at 10:39
Henno Brandsma
111k348119
edited Jan 20 at 10:39
Henno Brandsma
111k348119
edited Jan 20 at 10:39
Henno Brandsma
111k348119
111k348119
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
asked Jan 20 at 7:46
Josef KlimukJosef Klimuk
1205
1205
$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
1
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
1
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21
|
show 4 more comments
$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
1
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
1
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21
$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
1
1
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
1
1
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21
|
show 4 more comments
1 Answer
1
active
oldest
votes
1
$begingroup$
The powerset of $A$ is ${emptyset, A, {a}, {b}, {emptyset}, {a,b}, {a, emptyset}, {b, emptyset}}$, which has $8$ elements as $3$ has three.
And given $f$ the set $B$ always exists by the axioms of set theory, it is not ensured to be non-empty though, but the empty set does exist. We only need a set that is not in the image of $f$, and this set could well be $emptyset$, which is always a member of the powerset of any set.
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
|
show 6 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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1
$begingroup$
The powerset of $A$ is ${emptyset, A, {a}, {b}, {emptyset}, {a,b}, {a, emptyset}, {b, emptyset}}$, which has $8$ elements as $3$ has three.
And given $f$ the set $B$ always exists by the axioms of set theory, it is not ensured to be non-empty though, but the empty set does exist. We only need a set that is not in the image of $f$, and this set could well be $emptyset$, which is always a member of the powerset of any set.
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
|
show 6 more comments
1
$begingroup$
The powerset of $A$ is ${emptyset, A, {a}, {b}, {emptyset}, {a,b}, {a, emptyset}, {b, emptyset}}$, which has $8$ elements as $3$ has three.
And given $f$ the set $B$ always exists by the axioms of set theory, it is not ensured to be non-empty though, but the empty set does exist. We only need a set that is not in the image of $f$, and this set could well be $emptyset$, which is always a member of the powerset of any set.
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
|
show 6 more comments
1
1
1
$begingroup$
The powerset of $A$ is ${emptyset, A, {a}, {b}, {emptyset}, {a,b}, {a, emptyset}, {b, emptyset}}$, which has $8$ elements as $3$ has three.
And given $f$ the set $B$ always exists by the axioms of set theory, it is not ensured to be non-empty though, but the empty set does exist. We only need a set that is not in the image of $f$, and this set could well be $emptyset$, which is always a member of the powerset of any set.
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
The powerset of $A$ is ${emptyset, A, {a}, {b}, {emptyset}, {a,b}, {a, emptyset}, {b, emptyset}}$, which has $8$ elements as $3$ has three.
And given $f$ the set $B$ always exists by the axioms of set theory, it is not ensured to be non-empty though, but the empty set does exist. We only need a set that is not in the image of $f$, and this set could well be $emptyset$, which is always a member of the powerset of any set.
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
edited Jan 20 at 10:33
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
answered Jan 20 at 8:03
Henno BrandsmaHenno Brandsma
111k348119
111k348119
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
|
show 6 more comments
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
Since ∅ is a subset of every set, A={a,b} also contains ∅. So, P(A)={{a}, {b},{a,∅},{b,∅}, {a,b},{a,b,∅},{∅}}. So, P(A) will have 7 members, so the rule: P(A)=2^A will not work.
$endgroup$
– Josef Klimuk
Jan 21 at 5:39
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
@JosefKlimuk $emptyset$ is indeed a subset of each set, so an element of each powerset. You are removing it as an element. If $A={a,b}$ it’s subsets are $emptyset, {a},,{b},A$, so four subsets. You have some misconceptions about the powerset.
$endgroup$
– Henno Brandsma
Jan 21 at 5:44
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
What does it mean: "You are removing it as an element"? ∅ will be always in A, so will take part in formation of all A's subsets, which is P(A)
$endgroup$
– Josef Klimuk
Jan 21 at 6:02
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
$begingroup$
@JosefKlimuk in your $P(A)$ (comment above) you don’t mention $emptyset$ as an element.
$endgroup$
– Henno Brandsma
Jan 21 at 6:03
1
1
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
$begingroup$
No, you misunderstand what a correspondence is. $emptyset$ is not an element of $A$, so it has no "image". A correspondence from $A$ to $B$ is a set of pairs $(x,y)$ where for each $x in A$ there is a unique $y in B$ so that $(x,y)$ is in that correspondence. Here we have no elements in $A$, so the correspondence set is empty. So it has no sets in its image either.
$endgroup$
– Henno Brandsma
Jan 21 at 6:43
|
show 6 more comments
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$begingroup$
If $A$ has three elements, then $mathcal P(A)$ must have $2^3=8$ elements.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 7:52
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}, -⟷{∅}, -⟷{a,b}
$endgroup$
– Josef Klimuk
Jan 20 at 8:02
$begingroup$
A={a,b} P(A)={{a},{b},{a,b},{∅}} Possible correspondence: a⟷{a}, b⟷{b}. So, there is no such that B={x∈A|x∉f(x)}
$endgroup$
– Josef Klimuk
Jan 20 at 8:22
1
$begingroup$
Also in your post example, $B=emptyset$, also not in the image of the correspondence. The proof works.
$endgroup$
– Henno Brandsma
Jan 20 at 10:41
1
$begingroup$
In that example there does not exist $x$ such that $xnotin f(x)$. That does not say $B$ does not exist, it just says $B$ is empty.
$endgroup$
– David C. Ullrich
Jan 20 at 18:21