Zorn's Lemma implies Axiom of Choice
Zorn's Lemma implies Axiom of Choice
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $S$ be a collection of nonempty sets and $F$ be the collection of all functions $f$ for which ${rm dom}(f) subseteq S$ and $f(X)in X$ for all $X in {rm dom}(f)$. The set $F$ is ordered by inclusion $subseteq$.
Assume that $C$ is a chain in $(F,subseteq)$. Let $f_0=bigcup C$. It is easy to verify that $f_0 in F$ and $f_0$ is an upper bound of $C$. Then $F$ has a maximal element $bar f$ by Zorn's Lemma. To accomplish the proof, we next show that ${rm dom}(bar f)=S$. If not, ${rm dom}(bar f) subsetneq S$ and thus $S setminus {rm dom}(bar f) neq emptyset$. Take some $bar X in S setminus {rm dom}(bar f)$ and $bar xin bar X$. Clearly, $bar f bigcup {(bar X,bar x)}in F$. This contradicts the minimality of $bar f$. This completes the proof.
proof-verification elementary-set-theory axiom-of-choice
edited Dec 7 at 6:37
Shaun
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
add a comment |
Zorn's Lemma implies Axiom of Choice
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $S$ be a collection of nonempty sets and $F$ be the collection of all functions $f$ for which ${rm dom}(f) subseteq S$ and $f(X)in X$ for all $X in {rm dom}(f)$. The set $F$ is ordered by inclusion $subseteq$.
Assume that $C$ is a chain in $(F,subseteq)$. Let $f_0=bigcup C$. It is easy to verify that $f_0 in F$ and $f_0$ is an upper bound of $C$. Then $F$ has a maximal element $bar f$ by Zorn's Lemma. To accomplish the proof, we next show that ${rm dom}(bar f)=S$. If not, ${rm dom}(bar f) subsetneq S$ and thus $S setminus {rm dom}(bar f) neq emptyset$. Take some $bar X in S setminus {rm dom}(bar f)$ and $bar xin bar X$. Clearly, $bar f bigcup {(bar X,bar x)}in F$. This contradicts the minimality of $bar f$. This completes the proof.
proof-verification elementary-set-theory axiom-of-choice
edited Dec 7 at 6:37
Shaun
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15
add a comment |
Zorn's Lemma implies Axiom of Choice
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $S$ be a collection of nonempty sets and $F$ be the collection of all functions $f$ for which ${rm dom}(f) subseteq S$ and $f(X)in X$ for all $X in {rm dom}(f)$. The set $F$ is ordered by inclusion $subseteq$.
Assume that $C$ is a chain in $(F,subseteq)$. Let $f_0=bigcup C$. It is easy to verify that $f_0 in F$ and $f_0$ is an upper bound of $C$. Then $F$ has a maximal element $bar f$ by Zorn's Lemma. To accomplish the proof, we next show that ${rm dom}(bar f)=S$. If not, ${rm dom}(bar f) subsetneq S$ and thus $S setminus {rm dom}(bar f) neq emptyset$. Take some $bar X in S setminus {rm dom}(bar f)$ and $bar xin bar X$. Clearly, $bar f bigcup {(bar X,bar x)}in F$. This contradicts the minimality of $bar f$. This completes the proof.
proof-verification elementary-set-theory axiom-of-choice
edited Dec 7 at 6:37
Shaun
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
Zorn's Lemma implies Axiom of Choice
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $S$ be a collection of nonempty sets and $F$ be the collection of all functions $f$ for which ${rm dom}(f) subseteq S$ and $f(X)in X$ for all $X in {rm dom}(f)$. The set $F$ is ordered by inclusion $subseteq$.
Assume that $C$ is a chain in $(F,subseteq)$. Let $f_0=bigcup C$. It is easy to verify that $f_0 in F$ and $f_0$ is an upper bound of $C$. Then $F$ has a maximal element $bar f$ by Zorn's Lemma. To accomplish the proof, we next show that ${rm dom}(bar f)=S$. If not, ${rm dom}(bar f) subsetneq S$ and thus $S setminus {rm dom}(bar f) neq emptyset$. Take some $bar X in S setminus {rm dom}(bar f)$ and $bar xin bar X$. Clearly, $bar f bigcup {(bar X,bar x)}in F$. This contradicts the minimality of $bar f$. This completes the proof.
proof-verification elementary-set-theory axiom-of-choice
proof-verification elementary-set-theory axiom-of-choice
edited Dec 7 at 6:37
Shaun
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
edited Dec 7 at 6:37
Shaun
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
edited Dec 7 at 6:37
Shaun
8,344113578
edited Dec 7 at 6:37
Shaun
8,344113578
edited Dec 7 at 6:37
Shaun
8,344113578
8,344113578
asked Dec 7 at 6:21
Le Anh Dung
9261421
asked Dec 7 at 6:21
Le Anh Dung
9261421
asked Dec 7 at 6:21
Le Anh Dung
9261421
9261421
math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15
add a comment |
math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15
math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15
math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15
add a comment |
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math.stackexchange.com/questions/2088040/…
– Asaf Karagila♦
Dec 7 at 7:15