Question about set order theory
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Suppose we have an ordered set $(A,le)$ with the property that, for any decreasing chain (sequence) $a_0 ge a_1 ge dotsb$, there exists $n in mathbb{N}$ such that, for each $k in mathbb{N}$, we have $a_n=a_{n+k}$.
We have to prove that any non empty subset $B subseteq A$ has a minimal element.
I'm new to order theory, so I can not figure out how to deal with this problem.
elementary-set-theory relations order-theory recursion
edited Nov 26 at 11:28
egreg
176k1384198
asked Nov 26 at 10:40
Raul1998
63
add a comment |
up vote
0
down vote
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Suppose we have an ordered set $(A,le)$ with the property that, for any decreasing chain (sequence) $a_0 ge a_1 ge dotsb$, there exists $n in mathbb{N}$ such that, for each $k in mathbb{N}$, we have $a_n=a_{n+k}$.
We have to prove that any non empty subset $B subseteq A$ has a minimal element.
I'm new to order theory, so I can not figure out how to deal with this problem.
elementary-set-theory relations order-theory recursion
edited Nov 26 at 11:28
egreg
176k1384198
asked Nov 26 at 10:40
Raul1998
63
1
What has your question to do withlinear-algebra,abstract-algebra, ormultisets?
– José Carlos Santos
Nov 26 at 10:42
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have an ordered set $(A,le)$ with the property that, for any decreasing chain (sequence) $a_0 ge a_1 ge dotsb$, there exists $n in mathbb{N}$ such that, for each $k in mathbb{N}$, we have $a_n=a_{n+k}$.
We have to prove that any non empty subset $B subseteq A$ has a minimal element.
I'm new to order theory, so I can not figure out how to deal with this problem.
elementary-set-theory relations order-theory recursion
edited Nov 26 at 11:28
egreg
176k1384198
asked Nov 26 at 10:40
Raul1998
63
Suppose we have an ordered set $(A,le)$ with the property that, for any decreasing chain (sequence) $a_0 ge a_1 ge dotsb$, there exists $n in mathbb{N}$ such that, for each $k in mathbb{N}$, we have $a_n=a_{n+k}$.
We have to prove that any non empty subset $B subseteq A$ has a minimal element.
I'm new to order theory, so I can not figure out how to deal with this problem.
elementary-set-theory relations order-theory recursion
elementary-set-theory relations order-theory recursion
edited Nov 26 at 11:28
egreg
176k1384198
asked Nov 26 at 10:40
Raul1998
63
edited Nov 26 at 11:28
egreg
176k1384198
asked Nov 26 at 10:40
Raul1998
63
edited Nov 26 at 11:28
egreg
176k1384198
edited Nov 26 at 11:28
egreg
176k1384198
edited Nov 26 at 11:28
egreg
176k1384198
176k1384198
asked Nov 26 at 10:40
Raul1998
63
asked Nov 26 at 10:40
Raul1998
63
asked Nov 26 at 10:40
Raul1998
63
63
1
What has your question to do withlinear-algebra,abstract-algebra, ormultisets?
– José Carlos Santos
Nov 26 at 10:42
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48
add a comment |
1
What has your question to do withlinear-algebra,abstract-algebra, ormultisets?
– José Carlos Santos
Nov 26 at 10:42
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48
1
1
What has your question to do with
linear-algebra, abstract-algebra, or multisets?– José Carlos Santos
Nov 26 at 10:42
What has your question to do with
linear-algebra, abstract-algebra, or multisets?– José Carlos Santos
Nov 26 at 10:42
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48
add a comment |
1 Answer
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1
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You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $Bne varnothing$ does not possess a minimal element. Picking an $a_0in B$ you can define recursively the sequence
$$
text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}.
$$
Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element,
and this leads to a contradiction.
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $Bne varnothing$ does not possess a minimal element. Picking an $a_0in B$ you can define recursively the sequence
$$
text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}.
$$
Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element,
and this leads to a contradiction.
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
add a comment |
up vote
1
down vote
You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $Bne varnothing$ does not possess a minimal element. Picking an $a_0in B$ you can define recursively the sequence
$$
text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}.
$$
Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element,
and this leads to a contradiction.
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
add a comment |
up vote
1
down vote
up vote
1
down vote
You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $Bne varnothing$ does not possess a minimal element. Picking an $a_0in B$ you can define recursively the sequence
$$
text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}.
$$
Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element,
and this leads to a contradiction.
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $Bne varnothing$ does not possess a minimal element. Picking an $a_0in B$ you can define recursively the sequence
$$
text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}.
$$
Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element,
and this leads to a contradiction.
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 26 at 10:54
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
add a comment |
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
1
1
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
This is not really a proof by contradiction, but rather by contrapositive.
– egreg
Nov 26 at 11:30
add a comment |
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1
What has your question to do with
linear-algebra,abstract-algebra, ormultisets?– José Carlos Santos
Nov 26 at 10:42
@JoséCarlosSantos, sorry for that, fixed this.
– Raul1998
Nov 26 at 10:48