Prob. 8 (b), Sec. 10, in Munkres' TOPOLOGY, 2nd ed: The union of any collection of disjoint well-ordered sets...












1












$begingroup$


Here is Prob. 8, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:



Problem 8 (a):




Let $A_1$ and $A_2$ be disjoint sets, well-ordered by $<_1$ and $<_2$, respectively. Define an order relation on $A_1 cup A_2$ by letting $a < b$ either if $a, b in A_1$ and $a <_1 b$, or if $a, b in A_2$ and $a <_2 b$, or if $a in A_1$ and $b in A_2$. Show that this is a well-ordering.




Problem 8 (b):




Generalize (a) to an arbitrary family of disjoint well-ordered sets, indexed by a well-ordered set.




I think I'm clear as to the proof required in Prob. 8 (a).



My Attempt at Prob. 8 (b):




Let $J$ be a (non-empty) well ordered set; let $$left{ A_alpha colon alpha in J right}$$ be a collection of non-empty, (pairwise) disjoint well-ordered sets indexed by set $J$; and let $$ A colon= bigcup_{alpha in J} A_alpha. $$
For each $alpha in J$, let $<_alpha$ denote the well-ordering relation on the set $A_alpha$.



For any two elements $a, b in A$, let us define $a < b$ to mean the following:



Either $a, b in A_alpha$ for some $alpha in J$ and $a <_alpha b$, or $a in A_alpha$, $b in A_beta$ for some $alpha, beta in J$ such that $alpha <_J beta$.



Then the set $A$ is a well-ordered set.




Is this statement correct?



If so, then here is my proof:




Let $S$ be a non-empty subset of the set $A = bigcup_{alpha in J} A_alpha$. Let $J_0$ be the following subset of $J$.
$$ J_0 colon= left{ alpha in J colon S cap A_alpha neq emptyset right}. $$
Then the set $J_0$ is non-empty, and as such it has a smallest element, say $alpha_0$. Then $alpha_0 in J$ and $S cap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$ and so has a smallest element $a_{alpha_0}$, which is also the smallest of $S$ with respect to the order relation on $A$.




Is this proof correct? If so, then is my presentation accessible enough? If not, then where have I erred?






elementary-set-theory proof-verification logic order-theory well-orders







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
    $endgroup$
    – drhab
    Jan 1 '18 at 11:39
















1












$begingroup$


Here is Prob. 8, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:



Problem 8 (a):




Let $A_1$ and $A_2$ be disjoint sets, well-ordered by $<_1$ and $<_2$, respectively. Define an order relation on $A_1 cup A_2$ by letting $a < b$ either if $a, b in A_1$ and $a <_1 b$, or if $a, b in A_2$ and $a <_2 b$, or if $a in A_1$ and $b in A_2$. Show that this is a well-ordering.




Problem 8 (b):




Generalize (a) to an arbitrary family of disjoint well-ordered sets, indexed by a well-ordered set.




I think I'm clear as to the proof required in Prob. 8 (a).



My Attempt at Prob. 8 (b):




Let $J$ be a (non-empty) well ordered set; let $$left{ A_alpha colon alpha in J right}$$ be a collection of non-empty, (pairwise) disjoint well-ordered sets indexed by set $J$; and let $$ A colon= bigcup_{alpha in J} A_alpha. $$
For each $alpha in J$, let $<_alpha$ denote the well-ordering relation on the set $A_alpha$.



For any two elements $a, b in A$, let us define $a < b$ to mean the following:



Either $a, b in A_alpha$ for some $alpha in J$ and $a <_alpha b$, or $a in A_alpha$, $b in A_beta$ for some $alpha, beta in J$ such that $alpha <_J beta$.



Then the set $A$ is a well-ordered set.




Is this statement correct?



If so, then here is my proof:




Let $S$ be a non-empty subset of the set $A = bigcup_{alpha in J} A_alpha$. Let $J_0$ be the following subset of $J$.
$$ J_0 colon= left{ alpha in J colon S cap A_alpha neq emptyset right}. $$
Then the set $J_0$ is non-empty, and as such it has a smallest element, say $alpha_0$. Then $alpha_0 in J$ and $S cap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$ and so has a smallest element $a_{alpha_0}$, which is also the smallest of $S$ with respect to the order relation on $A$.




Is this proof correct? If so, then is my presentation accessible enough? If not, then where have I erred?






elementary-set-theory proof-verification logic order-theory well-orders







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
    $endgroup$
    – drhab
    Jan 1 '18 at 11:39














1












1








1





$begingroup$


Here is Prob. 8, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:



Problem 8 (a):




Let $A_1$ and $A_2$ be disjoint sets, well-ordered by $<_1$ and $<_2$, respectively. Define an order relation on $A_1 cup A_2$ by letting $a < b$ either if $a, b in A_1$ and $a <_1 b$, or if $a, b in A_2$ and $a <_2 b$, or if $a in A_1$ and $b in A_2$. Show that this is a well-ordering.




Problem 8 (b):




Generalize (a) to an arbitrary family of disjoint well-ordered sets, indexed by a well-ordered set.




I think I'm clear as to the proof required in Prob. 8 (a).



My Attempt at Prob. 8 (b):




Let $J$ be a (non-empty) well ordered set; let $$left{ A_alpha colon alpha in J right}$$ be a collection of non-empty, (pairwise) disjoint well-ordered sets indexed by set $J$; and let $$ A colon= bigcup_{alpha in J} A_alpha. $$
For each $alpha in J$, let $<_alpha$ denote the well-ordering relation on the set $A_alpha$.



For any two elements $a, b in A$, let us define $a < b$ to mean the following:



Either $a, b in A_alpha$ for some $alpha in J$ and $a <_alpha b$, or $a in A_alpha$, $b in A_beta$ for some $alpha, beta in J$ such that $alpha <_J beta$.



Then the set $A$ is a well-ordered set.




Is this statement correct?



If so, then here is my proof:




Let $S$ be a non-empty subset of the set $A = bigcup_{alpha in J} A_alpha$. Let $J_0$ be the following subset of $J$.
$$ J_0 colon= left{ alpha in J colon S cap A_alpha neq emptyset right}. $$
Then the set $J_0$ is non-empty, and as such it has a smallest element, say $alpha_0$. Then $alpha_0 in J$ and $S cap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$ and so has a smallest element $a_{alpha_0}$, which is also the smallest of $S$ with respect to the order relation on $A$.




Is this proof correct? If so, then is my presentation accessible enough? If not, then where have I erred?






elementary-set-theory proof-verification logic order-theory well-orders







share|cite|improve this question









$endgroup$




Here is Prob. 8, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:



Problem 8 (a):




Let $A_1$ and $A_2$ be disjoint sets, well-ordered by $<_1$ and $<_2$, respectively. Define an order relation on $A_1 cup A_2$ by letting $a < b$ either if $a, b in A_1$ and $a <_1 b$, or if $a, b in A_2$ and $a <_2 b$, or if $a in A_1$ and $b in A_2$. Show that this is a well-ordering.




Problem 8 (b):




Generalize (a) to an arbitrary family of disjoint well-ordered sets, indexed by a well-ordered set.




I think I'm clear as to the proof required in Prob. 8 (a).



My Attempt at Prob. 8 (b):




Let $J$ be a (non-empty) well ordered set; let $$left{ A_alpha colon alpha in J right}$$ be a collection of non-empty, (pairwise) disjoint well-ordered sets indexed by set $J$; and let $$ A colon= bigcup_{alpha in J} A_alpha. $$
For each $alpha in J$, let $<_alpha$ denote the well-ordering relation on the set $A_alpha$.



For any two elements $a, b in A$, let us define $a < b$ to mean the following:



Either $a, b in A_alpha$ for some $alpha in J$ and $a <_alpha b$, or $a in A_alpha$, $b in A_beta$ for some $alpha, beta in J$ such that $alpha <_J beta$.



Then the set $A$ is a well-ordered set.




Is this statement correct?



If so, then here is my proof:




Let $S$ be a non-empty subset of the set $A = bigcup_{alpha in J} A_alpha$. Let $J_0$ be the following subset of $J$.
$$ J_0 colon= left{ alpha in J colon S cap A_alpha neq emptyset right}. $$
Then the set $J_0$ is non-empty, and as such it has a smallest element, say $alpha_0$. Then $alpha_0 in J$ and $S cap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$ and so has a smallest element $a_{alpha_0}$, which is also the smallest of $S$ with respect to the order relation on $A$.




Is this proof correct? If so, then is my presentation accessible enough? If not, then where have I erred?






elementary-set-theory proof-verification logic order-theory well-orders




elementary-set-theory proof-verification logic order-theory well-orders






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 1 '18 at 11:29









Saaqib MahmoodSaaqib Mahmood

7,85542480




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  • 3




    $begingroup$
    The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
    $endgroup$
    – drhab
    Jan 1 '18 at 11:39














  • 3




    $begingroup$
    The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
    $endgroup$
    – drhab
    Jan 1 '18 at 11:39








3




3




$begingroup$
The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
$endgroup$
– drhab
Jan 1 '18 at 11:39




$begingroup$
The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $sin S$ and prove that indeed $a_{alpha_0}leq s$.
$endgroup$
– drhab
Jan 1 '18 at 11:39










1 Answer
1






active

oldest

votes


















0












$begingroup$

Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:




Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.







share|cite|improve this answer











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  • $begingroup$
    This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
    $endgroup$
    – aleph_two
    Dec 24 '18 at 4:18











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

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0












$begingroup$

Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:




Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
    $endgroup$
    – aleph_two
    Dec 24 '18 at 4:18
















0












$begingroup$

Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:




Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
    $endgroup$
    – aleph_two
    Dec 24 '18 at 4:18














0












0








0





$begingroup$

Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:




Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.







share|cite|improve this answer











$endgroup$



Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:




Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.








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share|cite|improve this answer



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answered Dec 24 '18 at 4:17


























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aleph_two













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    This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
    $endgroup$
    – aleph_two
    Dec 24 '18 at 4:18


















  • $begingroup$
    This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
    $endgroup$
    – aleph_two
    Dec 24 '18 at 4:18
















$begingroup$
This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 24 '18 at 4:18




$begingroup$
This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process.
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Dec 24 '18 at 4:18


















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