Mixing
Let $X$ and $Y$ be set, then $|X|le |Y|$ or $|Y|le|X|.$
$begingroup$
Theorem. Let $X$ and $Y$ be set, then $|X|le |Y|$ or $|Y|le|X|.$
Proof. We consider the family $$mathcal{F}=left{(A,f);middle |; Asubset X,;fcolon Ato Y;text{injective}right}.$$
We define a order relation on $mathcal{F}$ in the follow way
$$(A,f)le(B,g)quadtext{if and only if}quad Asubseteq B;text{and};g_{|A}=f.$$
Question 1. Why does $A$ have to be a proper subset of $X$?
Let $$C={(A_i,f_i),; iin I}$$
a chain in $mathcal{F}$. This means that if $i,jin I$, then $A_isubseteq A_j$ $big(f_{j|A_i}=f_ibig)$ or $A_jsubseteq A_i$ $big(f_{i|A_i}=f_jbig).$
We observe that if $ain A_i$ and $bin A_j$, then $a, bin A_j$ if $A_isubseteq A_j$ or $a,bin A_i$ if $A_jsubseteq A_i.$ We consider
$$A:=bigcup_{iin I} A_i,$$
and we define $fcolon Ato Y$ such that $f(a):=f_i(a)$ if $ain A_i.$ We observe that the application $f$ is well defined. In fact if $ain A_icap A_j$, then $ain A_i$ and $ain A_j$, therefore $f(a):=f_i(a)$ and $f(a):=f_j(a)$, however we can suppose that $A_isubseteq A_j$, then
$$f(a):=f_{j}(a)=f_{j|A_i}(a)=f_i(a).$$
Moreover $f$ is injective: let $a,bin A$ with $f(a)=f(b)$, then for what they said before they belong to the same $A_i$ for same $iin I$, thus
$$f_i(a)=f(a)=f(b)=f_i(b),$$
then $$f_i(a)=f_i(b),$$
therefore $a=b$ because $f_i$ is injective. We have proved that $(A,f)inmathcal{F}.$. Now, for definition $A$ is an upper bound of $mathcal{F}$, then by Zorn's Lemma $mathcal{F}$ admtis maximal elements. Let $(S,f)inmathcal{F}$ such that $Ane X$ and $f(S)ne Y$
Question 2 Why do we need to take $Sne X$ and $f(S)ne Y$?
We take $x_0in Xsetminus S$ and $y_0in Ysetminus f(S)$, we can exstends $f$ to $Scup {x_0}$ defining $f(x_0):=y_0.$ Therefore, if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y$.
Question 3. Why if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y?$
Thanks for patience!
set-theory proof-explanation
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
$endgroup$
add a comment |
$begingroup$
Theorem. Let $X$ and $Y$ be set, then $|X|le |Y|$ or $|Y|le|X|.$
Proof. We consider the family $$mathcal{F}=left{(A,f);middle |; Asubset X,;fcolon Ato Y;text{injective}right}.$$
We define a order relation on $mathcal{F}$ in the follow way
$$(A,f)le(B,g)quadtext{if and only if}quad Asubseteq B;text{and};g_{|A}=f.$$
Question 1. Why does $A$ have to be a proper subset of $X$?
Let $$C={(A_i,f_i),; iin I}$$
a chain in $mathcal{F}$. This means that if $i,jin I$, then $A_isubseteq A_j$ $big(f_{j|A_i}=f_ibig)$ or $A_jsubseteq A_i$ $big(f_{i|A_i}=f_jbig).$
We observe that if $ain A_i$ and $bin A_j$, then $a, bin A_j$ if $A_isubseteq A_j$ or $a,bin A_i$ if $A_jsubseteq A_i.$ We consider
$$A:=bigcup_{iin I} A_i,$$
and we define $fcolon Ato Y$ such that $f(a):=f_i(a)$ if $ain A_i.$ We observe that the application $f$ is well defined. In fact if $ain A_icap A_j$, then $ain A_i$ and $ain A_j$, therefore $f(a):=f_i(a)$ and $f(a):=f_j(a)$, however we can suppose that $A_isubseteq A_j$, then
$$f(a):=f_{j}(a)=f_{j|A_i}(a)=f_i(a).$$
Moreover $f$ is injective: let $a,bin A$ with $f(a)=f(b)$, then for what they said before they belong to the same $A_i$ for same $iin I$, thus
$$f_i(a)=f(a)=f(b)=f_i(b),$$
then $$f_i(a)=f_i(b),$$
therefore $a=b$ because $f_i$ is injective. We have proved that $(A,f)inmathcal{F}.$. Now, for definition $A$ is an upper bound of $mathcal{F}$, then by Zorn's Lemma $mathcal{F}$ admtis maximal elements. Let $(S,f)inmathcal{F}$ such that $Ane X$ and $f(S)ne Y$
Question 2 Why do we need to take $Sne X$ and $f(S)ne Y$?
We take $x_0in Xsetminus S$ and $y_0in Ysetminus f(S)$, we can exstends $f$ to $Scup {x_0}$ defining $f(x_0):=y_0.$ Therefore, if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y$.
Question 3. Why if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y?$
Thanks for patience!
set-theory proof-explanation
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
$endgroup$
$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
$begingroup$
Proof explanation question! Sorry.
$endgroup$
– Jack J.
Jan 12 at 17:46
$begingroup$
Then why is this tagged as proof verification and proof writing?
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:46
$begingroup$
I just corrected.
$endgroup$
– Jack J.
Jan 12 at 18:59
$begingroup$
I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
$endgroup$
– Asaf Karagila♦
Jan 13 at 1:10
add a comment |
$begingroup$
Theorem. Let $X$ and $Y$ be set, then $|X|le |Y|$ or $|Y|le|X|.$
Proof. We consider the family $$mathcal{F}=left{(A,f);middle |; Asubset X,;fcolon Ato Y;text{injective}right}.$$
We define a order relation on $mathcal{F}$ in the follow way
$$(A,f)le(B,g)quadtext{if and only if}quad Asubseteq B;text{and};g_{|A}=f.$$
Question 1. Why does $A$ have to be a proper subset of $X$?
Let $$C={(A_i,f_i),; iin I}$$
a chain in $mathcal{F}$. This means that if $i,jin I$, then $A_isubseteq A_j$ $big(f_{j|A_i}=f_ibig)$ or $A_jsubseteq A_i$ $big(f_{i|A_i}=f_jbig).$
We observe that if $ain A_i$ and $bin A_j$, then $a, bin A_j$ if $A_isubseteq A_j$ or $a,bin A_i$ if $A_jsubseteq A_i.$ We consider
$$A:=bigcup_{iin I} A_i,$$
and we define $fcolon Ato Y$ such that $f(a):=f_i(a)$ if $ain A_i.$ We observe that the application $f$ is well defined. In fact if $ain A_icap A_j$, then $ain A_i$ and $ain A_j$, therefore $f(a):=f_i(a)$ and $f(a):=f_j(a)$, however we can suppose that $A_isubseteq A_j$, then
$$f(a):=f_{j}(a)=f_{j|A_i}(a)=f_i(a).$$
Moreover $f$ is injective: let $a,bin A$ with $f(a)=f(b)$, then for what they said before they belong to the same $A_i$ for same $iin I$, thus
$$f_i(a)=f(a)=f(b)=f_i(b),$$
then $$f_i(a)=f_i(b),$$
therefore $a=b$ because $f_i$ is injective. We have proved that $(A,f)inmathcal{F}.$. Now, for definition $A$ is an upper bound of $mathcal{F}$, then by Zorn's Lemma $mathcal{F}$ admtis maximal elements. Let $(S,f)inmathcal{F}$ such that $Ane X$ and $f(S)ne Y$
Question 2 Why do we need to take $Sne X$ and $f(S)ne Y$?
We take $x_0in Xsetminus S$ and $y_0in Ysetminus f(S)$, we can exstends $f$ to $Scup {x_0}$ defining $f(x_0):=y_0.$ Therefore, if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y$.
Question 3. Why if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y?$
Thanks for patience!
set-theory proof-explanation
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
$endgroup$
Theorem. Let $X$ and $Y$ be set, then $|X|le |Y|$ or $|Y|le|X|.$
Proof. We consider the family $$mathcal{F}=left{(A,f);middle |; Asubset X,;fcolon Ato Y;text{injective}right}.$$
We define a order relation on $mathcal{F}$ in the follow way
$$(A,f)le(B,g)quadtext{if and only if}quad Asubseteq B;text{and};g_{|A}=f.$$
Question 1. Why does $A$ have to be a proper subset of $X$?
Let $$C={(A_i,f_i),; iin I}$$
a chain in $mathcal{F}$. This means that if $i,jin I$, then $A_isubseteq A_j$ $big(f_{j|A_i}=f_ibig)$ or $A_jsubseteq A_i$ $big(f_{i|A_i}=f_jbig).$
We observe that if $ain A_i$ and $bin A_j$, then $a, bin A_j$ if $A_isubseteq A_j$ or $a,bin A_i$ if $A_jsubseteq A_i.$ We consider
$$A:=bigcup_{iin I} A_i,$$
and we define $fcolon Ato Y$ such that $f(a):=f_i(a)$ if $ain A_i.$ We observe that the application $f$ is well defined. In fact if $ain A_icap A_j$, then $ain A_i$ and $ain A_j$, therefore $f(a):=f_i(a)$ and $f(a):=f_j(a)$, however we can suppose that $A_isubseteq A_j$, then
$$f(a):=f_{j}(a)=f_{j|A_i}(a)=f_i(a).$$
Moreover $f$ is injective: let $a,bin A$ with $f(a)=f(b)$, then for what they said before they belong to the same $A_i$ for same $iin I$, thus
$$f_i(a)=f(a)=f(b)=f_i(b),$$
then $$f_i(a)=f_i(b),$$
therefore $a=b$ because $f_i$ is injective. We have proved that $(A,f)inmathcal{F}.$. Now, for definition $A$ is an upper bound of $mathcal{F}$, then by Zorn's Lemma $mathcal{F}$ admtis maximal elements. Let $(S,f)inmathcal{F}$ such that $Ane X$ and $f(S)ne Y$
Question 2 Why do we need to take $Sne X$ and $f(S)ne Y$?
We take $x_0in Xsetminus S$ and $y_0in Ysetminus f(S)$, we can exstends $f$ to $Scup {x_0}$ defining $f(x_0):=y_0.$ Therefore, if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y$.
Question 3. Why if $(S,f)$ is maximal in $mathcal{F}$, then $S=X$ or $f(S)=Y?$
Thanks for patience!
set-theory proof-explanation
set-theory proof-explanation
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
edited Jan 13 at 8:50
Jack J.
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
asked Jan 12 at 17:11
Jack J.Jack J.
4552419
4552419
$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
$begingroup$
Proof explanation question! Sorry.
$endgroup$
– Jack J.
Jan 12 at 17:46
$begingroup$
Then why is this tagged as proof verification and proof writing?
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:46
$begingroup$
I just corrected.
$endgroup$
– Jack J.
Jan 12 at 18:59
$begingroup$
I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
$endgroup$
– Asaf Karagila♦
Jan 13 at 1:10
add a comment |
$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
$begingroup$
Proof explanation question! Sorry.
$endgroup$
– Jack J.
Jan 12 at 17:46
$begingroup$
Then why is this tagged as proof verification and proof writing?
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:46
$begingroup$
I just corrected.
$endgroup$
– Jack J.
Jan 12 at 18:59
$begingroup$
I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
$endgroup$
– Asaf Karagila♦
Jan 13 at 1:10
$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
$begingroup$
Proof explanation question! Sorry.
$endgroup$
– Jack J.
Jan 12 at 17:46
$begingroup$
Proof explanation question! Sorry.
$endgroup$
– Jack J.
Jan 12 at 17:46
$begingroup$
Then why is this tagged as proof verification and proof writing?
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:46
$begingroup$
Then why is this tagged as proof verification and proof writing?
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:46
$begingroup$
I just corrected.
$endgroup$
– Jack J.
Jan 12 at 18:59
$begingroup$
I just corrected.
$endgroup$
– Jack J.
Jan 12 at 18:59
$begingroup$
I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
$endgroup$
– Asaf Karagila♦
Jan 13 at 1:10
$begingroup$
I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
$endgroup$
– Asaf Karagila♦
Jan 13 at 1:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the first question, this is a notational issue. Since you do not cite the source of the proof I can't give you an exact answer. However, the only way this makes sense is if $subset$ does not indicate proper subsets. Since this is something that some authors do, I'm inclined to believe this is the case here as well.
For the second and third questions, we don't need to take $A$ and $f(A)$ that way. We want to show that if $A$ is not $X$ and $f(A)$ is not $Y$, then $(A,f)$ is not a maximal element in the partial order. In particular, a maximal element—if it exists—must have either $A=X$ or $f(A)=Y$.
But finally, by Zorn's lemma, such maximal element does exist. So we are done.
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
$endgroup$
add a comment |
$begingroup$
Question 3: if both A $subsetneq$ X and f(A) $subsetneq$ Y, then the construction made a few lines above would be possible, thus extending ($A, f$) and getting to a contradiction (since ($A, f$) is maximal). I believe that the assumptions on which you have doubts (those of Question 2) are exactly those needed to extend a generic element ($S, f$).
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
For the first question, this is a notational issue. Since you do not cite the source of the proof I can't give you an exact answer. However, the only way this makes sense is if $subset$ does not indicate proper subsets. Since this is something that some authors do, I'm inclined to believe this is the case here as well.
For the second and third questions, we don't need to take $A$ and $f(A)$ that way. We want to show that if $A$ is not $X$ and $f(A)$ is not $Y$, then $(A,f)$ is not a maximal element in the partial order. In particular, a maximal element—if it exists—must have either $A=X$ or $f(A)=Y$.
But finally, by Zorn's lemma, such maximal element does exist. So we are done.
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
$endgroup$
add a comment |
$begingroup$
For the first question, this is a notational issue. Since you do not cite the source of the proof I can't give you an exact answer. However, the only way this makes sense is if $subset$ does not indicate proper subsets. Since this is something that some authors do, I'm inclined to believe this is the case here as well.
For the second and third questions, we don't need to take $A$ and $f(A)$ that way. We want to show that if $A$ is not $X$ and $f(A)$ is not $Y$, then $(A,f)$ is not a maximal element in the partial order. In particular, a maximal element—if it exists—must have either $A=X$ or $f(A)=Y$.
But finally, by Zorn's lemma, such maximal element does exist. So we are done.
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
$endgroup$
add a comment |
$begingroup$
For the first question, this is a notational issue. Since you do not cite the source of the proof I can't give you an exact answer. However, the only way this makes sense is if $subset$ does not indicate proper subsets. Since this is something that some authors do, I'm inclined to believe this is the case here as well.
For the second and third questions, we don't need to take $A$ and $f(A)$ that way. We want to show that if $A$ is not $X$ and $f(A)$ is not $Y$, then $(A,f)$ is not a maximal element in the partial order. In particular, a maximal element—if it exists—must have either $A=X$ or $f(A)=Y$.
But finally, by Zorn's lemma, such maximal element does exist. So we are done.
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
$endgroup$
For the first question, this is a notational issue. Since you do not cite the source of the proof I can't give you an exact answer. However, the only way this makes sense is if $subset$ does not indicate proper subsets. Since this is something that some authors do, I'm inclined to believe this is the case here as well.
For the second and third questions, we don't need to take $A$ and $f(A)$ that way. We want to show that if $A$ is not $X$ and $f(A)$ is not $Y$, then $(A,f)$ is not a maximal element in the partial order. In particular, a maximal element—if it exists—must have either $A=X$ or $f(A)=Y$.
But finally, by Zorn's lemma, such maximal element does exist. So we are done.
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
answered Jan 13 at 1:12
Asaf Karagila♦Asaf Karagila
304k32431763
304k32431763
add a comment |
add a comment |
$begingroup$
Question 3: if both A $subsetneq$ X and f(A) $subsetneq$ Y, then the construction made a few lines above would be possible, thus extending ($A, f$) and getting to a contradiction (since ($A, f$) is maximal). I believe that the assumptions on which you have doubts (those of Question 2) are exactly those needed to extend a generic element ($S, f$).
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
$endgroup$
add a comment |
$begingroup$
Question 3: if both A $subsetneq$ X and f(A) $subsetneq$ Y, then the construction made a few lines above would be possible, thus extending ($A, f$) and getting to a contradiction (since ($A, f$) is maximal). I believe that the assumptions on which you have doubts (those of Question 2) are exactly those needed to extend a generic element ($S, f$).
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
$endgroup$
add a comment |
$begingroup$
Question 3: if both A $subsetneq$ X and f(A) $subsetneq$ Y, then the construction made a few lines above would be possible, thus extending ($A, f$) and getting to a contradiction (since ($A, f$) is maximal). I believe that the assumptions on which you have doubts (those of Question 2) are exactly those needed to extend a generic element ($S, f$).
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
$endgroup$
Question 3: if both A $subsetneq$ X and f(A) $subsetneq$ Y, then the construction made a few lines above would be possible, thus extending ($A, f$) and getting to a contradiction (since ($A, f$) is maximal). I believe that the assumptions on which you have doubts (those of Question 2) are exactly those needed to extend a generic element ($S, f$).
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
answered Jan 12 at 17:51
Simone RamelloSimone Ramello
1478
1478
add a comment |
add a comment |
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$begingroup$
Is this a proof verification question or a proof explanation question? Also, note that $subset$ is used by some authors as $subseteq$ would be used by others.
$endgroup$
– Asaf Karagila♦
Jan 12 at 17:45
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Proof explanation question! Sorry.
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– Jack J.
Jan 12 at 17:46
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Then why is this tagged as proof verification and proof writing?
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– Asaf Karagila♦
Jan 12 at 17:46
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I just corrected.
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– Jack J.
Jan 12 at 18:59
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I think you have a typo there with $S$ and $f(S)$, both of which should have $A$ and $f(A)$ respectively.
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– Asaf Karagila♦
Jan 13 at 1:10