Mixing
Let $A$ be a chain. $B,C$ subsets of $A$ s.t $A= Bcup C$. If $B$ and $C$ are well-ordered then A is...
$begingroup$
My attempt:
Assume that $B$ and $C$ are well-ordered and A is not.
Since A is not well-ordered it contains an infinite descending sequence ${a_i}_{i=0}^infty$. If the sequence has a limit $l$ $in$ $A$ then $l$ is in B or C. With no loss of generality assume $l in B$, then we can find $N in mathbb{N}$ such that ${a_i}_{i=N}^infty$ $subseteq B$. Then $B$ contains an infinite descending sequence hence it is not well ordered and we have a contradiction.
Question: How can I treat the case ${a_i}_{i=0}^infty$ is divergent?
elementary-set-theory well-orders
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
$endgroup$
add a comment |
$begingroup$
My attempt:
Assume that $B$ and $C$ are well-ordered and A is not.
Since A is not well-ordered it contains an infinite descending sequence ${a_i}_{i=0}^infty$. If the sequence has a limit $l$ $in$ $A$ then $l$ is in B or C. With no loss of generality assume $l in B$, then we can find $N in mathbb{N}$ such that ${a_i}_{i=N}^infty$ $subseteq B$. Then $B$ contains an infinite descending sequence hence it is not well ordered and we have a contradiction.
Question: How can I treat the case ${a_i}_{i=0}^infty$ is divergent?
elementary-set-theory well-orders
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
$endgroup$
$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28
add a comment |
$begingroup$
My attempt:
Assume that $B$ and $C$ are well-ordered and A is not.
Since A is not well-ordered it contains an infinite descending sequence ${a_i}_{i=0}^infty$. If the sequence has a limit $l$ $in$ $A$ then $l$ is in B or C. With no loss of generality assume $l in B$, then we can find $N in mathbb{N}$ such that ${a_i}_{i=N}^infty$ $subseteq B$. Then $B$ contains an infinite descending sequence hence it is not well ordered and we have a contradiction.
Question: How can I treat the case ${a_i}_{i=0}^infty$ is divergent?
elementary-set-theory well-orders
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
$endgroup$
My attempt:
Assume that $B$ and $C$ are well-ordered and A is not.
Since A is not well-ordered it contains an infinite descending sequence ${a_i}_{i=0}^infty$. If the sequence has a limit $l$ $in$ $A$ then $l$ is in B or C. With no loss of generality assume $l in B$, then we can find $N in mathbb{N}$ such that ${a_i}_{i=N}^infty$ $subseteq B$. Then $B$ contains an infinite descending sequence hence it is not well ordered and we have a contradiction.
Question: How can I treat the case ${a_i}_{i=0}^infty$ is divergent?
elementary-set-theory well-orders
elementary-set-theory well-orders
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
edited Jan 27 at 14:27
José Carlos Santos
170k23132238
170k23132238
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
asked Jan 27 at 14:26
kolanamubeinnkolanamubeinn
112
112
$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28
add a comment |
$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28
$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28
$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Alternative.
Let $D$ be a non-empty subset of $A$.
Then $Dcap B$ is not empty or $Dcap C$ is not empty.
If both are not empty then let $b$ be smallest element of $Dcap B$ and let $c$ be smallest element of $Dcap C$.
$b$ and $c$ are comparable. If $bleq c$ then let $d=b$ and if $cleq b$ then let $d=c$.
Then it can be proved that $d$ is smallest element of $D$.
If only one of the sets $Dcap B$ and $Dcap C$ is not empty then let $d$ be the smallest element of the non-empty set.
Again it can be proved that $d$ is smallest element of $D$.
So $A$ is a totally ordered set such that every non-empty subset has a smallest element.
That means exactly that $A$ is well-ordered.
answered Jan 27 at 14:44
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
If $(a_n)_{ninBbb N}$ is a sequence, then either $B_a={xin {a_n}_{ninBbb N}mid xin B}$ or $C_a={xin {a_n}_{ninBbb N}mid xin C}$ are infinite, WOLG $B_a$ is infinite, then $n_k=min(minBbb Nmid a_min B_aland a_m <a_{n_{k-1}})=min(minBbb Nmid a_min B_aland m >n_{k-1})$, then $(a_{n_k})_{kinBbb N}$ is infinite decreasing sequence in $B$
answered Jan 27 at 14:33
HoloHolo
6,08421131
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Alternative.
Let $D$ be a non-empty subset of $A$.
Then $Dcap B$ is not empty or $Dcap C$ is not empty.
If both are not empty then let $b$ be smallest element of $Dcap B$ and let $c$ be smallest element of $Dcap C$.
$b$ and $c$ are comparable. If $bleq c$ then let $d=b$ and if $cleq b$ then let $d=c$.
Then it can be proved that $d$ is smallest element of $D$.
If only one of the sets $Dcap B$ and $Dcap C$ is not empty then let $d$ be the smallest element of the non-empty set.
Again it can be proved that $d$ is smallest element of $D$.
So $A$ is a totally ordered set such that every non-empty subset has a smallest element.
That means exactly that $A$ is well-ordered.
answered Jan 27 at 14:44
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Alternative.
Let $D$ be a non-empty subset of $A$.
Then $Dcap B$ is not empty or $Dcap C$ is not empty.
If both are not empty then let $b$ be smallest element of $Dcap B$ and let $c$ be smallest element of $Dcap C$.
$b$ and $c$ are comparable. If $bleq c$ then let $d=b$ and if $cleq b$ then let $d=c$.
Then it can be proved that $d$ is smallest element of $D$.
If only one of the sets $Dcap B$ and $Dcap C$ is not empty then let $d$ be the smallest element of the non-empty set.
Again it can be proved that $d$ is smallest element of $D$.
So $A$ is a totally ordered set such that every non-empty subset has a smallest element.
That means exactly that $A$ is well-ordered.
answered Jan 27 at 14:44
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Alternative.
Let $D$ be a non-empty subset of $A$.
Then $Dcap B$ is not empty or $Dcap C$ is not empty.
If both are not empty then let $b$ be smallest element of $Dcap B$ and let $c$ be smallest element of $Dcap C$.
$b$ and $c$ are comparable. If $bleq c$ then let $d=b$ and if $cleq b$ then let $d=c$.
Then it can be proved that $d$ is smallest element of $D$.
If only one of the sets $Dcap B$ and $Dcap C$ is not empty then let $d$ be the smallest element of the non-empty set.
Again it can be proved that $d$ is smallest element of $D$.
So $A$ is a totally ordered set such that every non-empty subset has a smallest element.
That means exactly that $A$ is well-ordered.
answered Jan 27 at 14:44
drhabdrhab
103k545136
$endgroup$
Alternative.
Let $D$ be a non-empty subset of $A$.
Then $Dcap B$ is not empty or $Dcap C$ is not empty.
If both are not empty then let $b$ be smallest element of $Dcap B$ and let $c$ be smallest element of $Dcap C$.
$b$ and $c$ are comparable. If $bleq c$ then let $d=b$ and if $cleq b$ then let $d=c$.
Then it can be proved that $d$ is smallest element of $D$.
If only one of the sets $Dcap B$ and $Dcap C$ is not empty then let $d$ be the smallest element of the non-empty set.
Again it can be proved that $d$ is smallest element of $D$.
So $A$ is a totally ordered set such that every non-empty subset has a smallest element.
That means exactly that $A$ is well-ordered.
answered Jan 27 at 14:44
drhabdrhab
103k545136
answered Jan 27 at 14:44
drhabdrhab
103k545136
answered Jan 27 at 14:44
drhabdrhab
103k545136
answered Jan 27 at 14:44
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
If $(a_n)_{ninBbb N}$ is a sequence, then either $B_a={xin {a_n}_{ninBbb N}mid xin B}$ or $C_a={xin {a_n}_{ninBbb N}mid xin C}$ are infinite, WOLG $B_a$ is infinite, then $n_k=min(minBbb Nmid a_min B_aland a_m <a_{n_{k-1}})=min(minBbb Nmid a_min B_aland m >n_{k-1})$, then $(a_{n_k})_{kinBbb N}$ is infinite decreasing sequence in $B$
answered Jan 27 at 14:33
HoloHolo
6,08421131
$endgroup$
add a comment |
$begingroup$
If $(a_n)_{ninBbb N}$ is a sequence, then either $B_a={xin {a_n}_{ninBbb N}mid xin B}$ or $C_a={xin {a_n}_{ninBbb N}mid xin C}$ are infinite, WOLG $B_a$ is infinite, then $n_k=min(minBbb Nmid a_min B_aland a_m <a_{n_{k-1}})=min(minBbb Nmid a_min B_aland m >n_{k-1})$, then $(a_{n_k})_{kinBbb N}$ is infinite decreasing sequence in $B$
answered Jan 27 at 14:33
HoloHolo
6,08421131
$endgroup$
add a comment |
$begingroup$
If $(a_n)_{ninBbb N}$ is a sequence, then either $B_a={xin {a_n}_{ninBbb N}mid xin B}$ or $C_a={xin {a_n}_{ninBbb N}mid xin C}$ are infinite, WOLG $B_a$ is infinite, then $n_k=min(minBbb Nmid a_min B_aland a_m <a_{n_{k-1}})=min(minBbb Nmid a_min B_aland m >n_{k-1})$, then $(a_{n_k})_{kinBbb N}$ is infinite decreasing sequence in $B$
answered Jan 27 at 14:33
HoloHolo
6,08421131
$endgroup$
If $(a_n)_{ninBbb N}$ is a sequence, then either $B_a={xin {a_n}_{ninBbb N}mid xin B}$ or $C_a={xin {a_n}_{ninBbb N}mid xin C}$ are infinite, WOLG $B_a$ is infinite, then $n_k=min(minBbb Nmid a_min B_aland a_m <a_{n_{k-1}})=min(minBbb Nmid a_min B_aland m >n_{k-1})$, then $(a_{n_k})_{kinBbb N}$ is infinite decreasing sequence in $B$
answered Jan 27 at 14:33
HoloHolo
6,08421131
edited Jan 27 at 14:40
answered Jan 27 at 14:33
HoloHolo
6,08421131
answered Jan 27 at 14:33
HoloHolo
6,08421131
answered Jan 27 at 14:33
HoloHolo
6,08421131
6,08421131
add a comment |
add a comment |
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$begingroup$
Hint: the sequence either has infinitely many elements in $B$ or in $C$.
$endgroup$
– Wojowu
Jan 27 at 14:28