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Prove that the set of all subsets of a set is itself a set.
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Problem from Tao's Analysis I.
Exercise 3.4.6. Prove Lemma 3.4.10. (Hint: start with the set $0,1^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^-1(1)$.)
I have read some answers on stackexchange ((exercise from Tao's analysis book) Proof of a lemma relating to power set of X , Finding a correspondence between $0,1^A$ and $mathcal P(A)$) but my proof is a little different and I want to know if it's correct and complete, using the provided axioms.
Lemma 3.4.10 Let $X$ be a set. Then the set $Y: Y textit is a subset of X $ is a set.
Axiom 3.7 (Replacement). Let $A$ be a set. For any object $xin A$, and any object $y$, suppose we have a statement $P(x,y)$
pertaining to $x$ and $y$, such that for each $x in A$ there at most
one $y$ for which $P(x,y)$ is true. Then there exists a set
$y:P(x,y)textit is true for some xin A$, such that for any
object $z$, $$ziny:P(x,y)textit is true for some xin A\ iff
> P(x,z) textit is true for some xin A$$
Axiom 3.11 (Power set axiom). Let $X$ and $Y$ be sets. Then there exists a set, denoted $Y^X$, which consits of all the functions from
$X$ to $Y$, thus $$fin Y^X iff textit ($f$ is a function with
domain $X$ and range $Y$).$$
proof. By Axiom 3.11, $0,1^X$ is a set, namely,
$$0,1^X=f: textit $f$ is a function from $X$ to $0,1$.$$
Then by Axiom 3.7,
$$A:A=f^-1(1 textit for some fin 0,1text is a set.$$
We will show that for all sets $A$, $Ain P$ if and only if $Asubseteq X$.
Let $Asubseteq X$, and let $I_A:Xrightarrow 0,1$ be the function such that
$$I_A(a)=begincases
0 text, if anotin A \
1 text, if ain A
endcases.$$
Since $I_A$ is a function from $X$ to $0,1$, $I_Ain 0,1^X$. We will show that $A=I_A^-1(1)$.
Suppose $ain A$. Then $I_A(a)=1$, hence $ain I_A^-1(1)$. Now suppose $ain I_A^-1(1)$. Then $I_A(a)in1$, hence $I_A(a)=1$, so $ain A$.
Thus $Ain P$ if and only if $Asubseteq X$, thus $P$ is the set of all subsets of $X$.
real-analysis proof-verification
asked Aug 2 at 22:54
Ken Tjhia
307
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Problem from Tao's Analysis I.
Exercise 3.4.6. Prove Lemma 3.4.10. (Hint: start with the set $0,1^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^-1(1)$.)
I have read some answers on stackexchange ((exercise from Tao's analysis book) Proof of a lemma relating to power set of X , Finding a correspondence between $0,1^A$ and $mathcal P(A)$) but my proof is a little different and I want to know if it's correct and complete, using the provided axioms.
Lemma 3.4.10 Let $X$ be a set. Then the set $Y: Y textit is a subset of X $ is a set.
Axiom 3.7 (Replacement). Let $A$ be a set. For any object $xin A$, and any object $y$, suppose we have a statement $P(x,y)$
pertaining to $x$ and $y$, such that for each $x in A$ there at most
one $y$ for which $P(x,y)$ is true. Then there exists a set
$y:P(x,y)textit is true for some xin A$, such that for any
object $z$, $$ziny:P(x,y)textit is true for some xin A\ iff
> P(x,z) textit is true for some xin A$$
Axiom 3.11 (Power set axiom). Let $X$ and $Y$ be sets. Then there exists a set, denoted $Y^X$, which consits of all the functions from
$X$ to $Y$, thus $$fin Y^X iff textit ($f$ is a function with
domain $X$ and range $Y$).$$
proof. By Axiom 3.11, $0,1^X$ is a set, namely,
$$0,1^X=f: textit $f$ is a function from $X$ to $0,1$.$$
Then by Axiom 3.7,
$$A:A=f^-1(1 textit for some fin 0,1text is a set.$$
We will show that for all sets $A$, $Ain P$ if and only if $Asubseteq X$.
Let $Asubseteq X$, and let $I_A:Xrightarrow 0,1$ be the function such that
$$I_A(a)=begincases
0 text, if anotin A \
1 text, if ain A
endcases.$$
Since $I_A$ is a function from $X$ to $0,1$, $I_Ain 0,1^X$. We will show that $A=I_A^-1(1)$.
Suppose $ain A$. Then $I_A(a)=1$, hence $ain I_A^-1(1)$. Now suppose $ain I_A^-1(1)$. Then $I_A(a)in1$, hence $I_A(a)=1$, so $ain A$.
Thus $Ain P$ if and only if $Asubseteq X$, thus $P$ is the set of all subsets of $X$.
real-analysis proof-verification
asked Aug 2 at 22:54
Ken Tjhia
307
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem from Tao's Analysis I.
Exercise 3.4.6. Prove Lemma 3.4.10. (Hint: start with the set $0,1^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^-1(1)$.)
I have read some answers on stackexchange ((exercise from Tao's analysis book) Proof of a lemma relating to power set of X , Finding a correspondence between $0,1^A$ and $mathcal P(A)$) but my proof is a little different and I want to know if it's correct and complete, using the provided axioms.
Lemma 3.4.10 Let $X$ be a set. Then the set $Y: Y textit is a subset of X $ is a set.
Axiom 3.7 (Replacement). Let $A$ be a set. For any object $xin A$, and any object $y$, suppose we have a statement $P(x,y)$
pertaining to $x$ and $y$, such that for each $x in A$ there at most
one $y$ for which $P(x,y)$ is true. Then there exists a set
$y:P(x,y)textit is true for some xin A$, such that for any
object $z$, $$ziny:P(x,y)textit is true for some xin A\ iff
> P(x,z) textit is true for some xin A$$
Axiom 3.11 (Power set axiom). Let $X$ and $Y$ be sets. Then there exists a set, denoted $Y^X$, which consits of all the functions from
$X$ to $Y$, thus $$fin Y^X iff textit ($f$ is a function with
domain $X$ and range $Y$).$$
proof. By Axiom 3.11, $0,1^X$ is a set, namely,
$$0,1^X=f: textit $f$ is a function from $X$ to $0,1$.$$
Then by Axiom 3.7,
$$A:A=f^-1(1 textit for some fin 0,1text is a set.$$
We will show that for all sets $A$, $Ain P$ if and only if $Asubseteq X$.
Let $Asubseteq X$, and let $I_A:Xrightarrow 0,1$ be the function such that
$$I_A(a)=begincases
0 text, if anotin A \
1 text, if ain A
endcases.$$
Since $I_A$ is a function from $X$ to $0,1$, $I_Ain 0,1^X$. We will show that $A=I_A^-1(1)$.
Suppose $ain A$. Then $I_A(a)=1$, hence $ain I_A^-1(1)$. Now suppose $ain I_A^-1(1)$. Then $I_A(a)in1$, hence $I_A(a)=1$, so $ain A$.
Thus $Ain P$ if and only if $Asubseteq X$, thus $P$ is the set of all subsets of $X$.
real-analysis proof-verification
asked Aug 2 at 22:54
Ken Tjhia
307
Problem from Tao's Analysis I.
Exercise 3.4.6. Prove Lemma 3.4.10. (Hint: start with the set $0,1^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^-1(1)$.)
I have read some answers on stackexchange ((exercise from Tao's analysis book) Proof of a lemma relating to power set of X , Finding a correspondence between $0,1^A$ and $mathcal P(A)$) but my proof is a little different and I want to know if it's correct and complete, using the provided axioms.
Lemma 3.4.10 Let $X$ be a set. Then the set $Y: Y textit is a subset of X $ is a set.
Axiom 3.7 (Replacement). Let $A$ be a set. For any object $xin A$, and any object $y$, suppose we have a statement $P(x,y)$
pertaining to $x$ and $y$, such that for each $x in A$ there at most
one $y$ for which $P(x,y)$ is true. Then there exists a set
$y:P(x,y)textit is true for some xin A$, such that for any
object $z$, $$ziny:P(x,y)textit is true for some xin A\ iff
> P(x,z) textit is true for some xin A$$
Axiom 3.11 (Power set axiom). Let $X$ and $Y$ be sets. Then there exists a set, denoted $Y^X$, which consits of all the functions from
$X$ to $Y$, thus $$fin Y^X iff textit ($f$ is a function with
domain $X$ and range $Y$).$$
proof. By Axiom 3.11, $0,1^X$ is a set, namely,
$$0,1^X=f: textit $f$ is a function from $X$ to $0,1$.$$
Then by Axiom 3.7,
$$A:A=f^-1(1 textit for some fin 0,1text is a set.$$
We will show that for all sets $A$, $Ain P$ if and only if $Asubseteq X$.
Let $Asubseteq X$, and let $I_A:Xrightarrow 0,1$ be the function such that
$$I_A(a)=begincases
0 text, if anotin A \
1 text, if ain A
endcases.$$
Since $I_A$ is a function from $X$ to $0,1$, $I_Ain 0,1^X$. We will show that $A=I_A^-1(1)$.
Suppose $ain A$. Then $I_A(a)=1$, hence $ain I_A^-1(1)$. Now suppose $ain I_A^-1(1)$. Then $I_A(a)in1$, hence $I_A(a)=1$, so $ain A$.
Thus $Ain P$ if and only if $Asubseteq X$, thus $P$ is the set of all subsets of $X$.
real-analysis proof-verification
asked Aug 2 at 22:54
Ken Tjhia
307
asked Aug 2 at 22:54
Ken Tjhia
307
asked Aug 2 at 22:54
Ken Tjhia
307
asked Aug 2 at 22:54
Ken Tjhia
307
307
add a comment |Â
add a comment |Â
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