Can we expand “induction principle” to a partial order $(X, leq)$?
$begingroup$
We know that every infinite can be made well-ordered with an unknown order.
Also we can expand the induction principle on any infinite set in
the sense that
it can made well ordered.
Now partially ordered set may not be a well ordered set with respect
to the partial order.
Let a partially ordered set $(X, leq)$ with respect to this particular
order $'leq'$ and suppose that this partial order $'leq '$ does not make the
set $X$ well-ordered.
My question is-
Can we expand "induction principle" to the partially order set $(X,leq)$ keeping
in mind that $(X,leq)$ is not well ordered?
I have great confusion here.
elementary-set-theory order-theory
edited Mar 20 at 13:44
user21820
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
|
show 1 more comment
$begingroup$
We know that every infinite can be made well-ordered with an unknown order.
Also we can expand the induction principle on any infinite set in
the sense that
it can made well ordered.
Now partially ordered set may not be a well ordered set with respect
to the partial order.
Let a partially ordered set $(X, leq)$ with respect to this particular
order $'leq'$ and suppose that this partial order $'leq '$ does not make the
set $X$ well-ordered.
My question is-
Can we expand "induction principle" to the partially order set $(X,leq)$ keeping
in mind that $(X,leq)$ is not well ordered?
I have great confusion here.
elementary-set-theory order-theory
edited Mar 20 at 13:44
user21820
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
$begingroup$
@MarcvanLeeuwen, I did not get you. Can you elaborate it?
$endgroup$
– M. A. SARKAR
Mar 21 at 4:26
$begingroup$
For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
$endgroup$
– Marc van Leeuwen
Mar 21 at 4:57
$begingroup$
@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
$endgroup$
– M. A. SARKAR
Mar 21 at 6:44
$begingroup$
Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41
|
show 1 more comment
$begingroup$
We know that every infinite can be made well-ordered with an unknown order.
Also we can expand the induction principle on any infinite set in
the sense that
it can made well ordered.
Now partially ordered set may not be a well ordered set with respect
to the partial order.
Let a partially ordered set $(X, leq)$ with respect to this particular
order $'leq'$ and suppose that this partial order $'leq '$ does not make the
set $X$ well-ordered.
My question is-
Can we expand "induction principle" to the partially order set $(X,leq)$ keeping
in mind that $(X,leq)$ is not well ordered?
I have great confusion here.
elementary-set-theory order-theory
edited Mar 20 at 13:44
user21820
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
We know that every infinite can be made well-ordered with an unknown order.
Also we can expand the induction principle on any infinite set in
the sense that
it can made well ordered.
Now partially ordered set may not be a well ordered set with respect
to the partial order.
Let a partially ordered set $(X, leq)$ with respect to this particular
order $'leq'$ and suppose that this partial order $'leq '$ does not make the
set $X$ well-ordered.
My question is-
Can we expand "induction principle" to the partially order set $(X,leq)$ keeping
in mind that $(X,leq)$ is not well ordered?
I have great confusion here.
elementary-set-theory order-theory
elementary-set-theory order-theory
edited Mar 20 at 13:44
user21820
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
edited Mar 20 at 13:44
user21820
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
edited Mar 20 at 13:44
user21820
39.8k544158
edited Mar 20 at 13:44
user21820
39.8k544158
edited Mar 20 at 13:44
user21820
39.8k544158
39.8k544158
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
asked Mar 20 at 8:36
M. A. SARKARM. A. SARKAR
2,4551820
2,4551820
$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
$begingroup$
@MarcvanLeeuwen, I did not get you. Can you elaborate it?
$endgroup$
– M. A. SARKAR
Mar 21 at 4:26
$begingroup$
For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
$endgroup$
– Marc van Leeuwen
Mar 21 at 4:57
$begingroup$
@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
$endgroup$
– M. A. SARKAR
Mar 21 at 6:44
$begingroup$
Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41
|
show 1 more comment
$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
$begingroup$
@MarcvanLeeuwen, I did not get you. Can you elaborate it?
$endgroup$
– M. A. SARKAR
Mar 21 at 4:26
$begingroup$
For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
$endgroup$
– Marc van Leeuwen
Mar 21 at 4:57
$begingroup$
@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
$endgroup$
– M. A. SARKAR
Mar 21 at 6:44
$begingroup$
Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41
$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
$begingroup$
@MarcvanLeeuwen, I did not get you. Can you elaborate it?
$endgroup$
– M. A. SARKAR
Mar 21 at 4:26
$begingroup$
@MarcvanLeeuwen, I did not get you. Can you elaborate it?
$endgroup$
– M. A. SARKAR
Mar 21 at 4:26
$begingroup$
For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
$endgroup$
– Marc van Leeuwen
Mar 21 at 4:57
$begingroup$
For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
$endgroup$
– Marc van Leeuwen
Mar 21 at 4:57
$begingroup$
@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
$endgroup$
– M. A. SARKAR
Mar 21 at 6:44
$begingroup$
@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
$endgroup$
– M. A. SARKAR
Mar 21 at 6:44
$begingroup$
Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41
$begingroup$
Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
We can't even do induction on a total order if it's not well-ordered. Like on $Bbb Q$ or $Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
answered Mar 20 at 8:56
ArthurArthur
121k7121207
$endgroup$
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
$begingroup$
@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
$endgroup$
– Arthur
Mar 20 at 13:54
|
show 3 more comments
$begingroup$
Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m in S$ such that there is no $x in S$ with $x R m$.
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set.
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
$endgroup$
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
3
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
add a comment |
$begingroup$
One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $xnotin S$ it is true that $forall x: (forall y: (y<xto P(y))to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can't even do induction on a total order if it's not well-ordered. Like on $Bbb Q$ or $Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
answered Mar 20 at 8:56
ArthurArthur
121k7121207
$endgroup$
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
$begingroup$
@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
$endgroup$
– Arthur
Mar 20 at 13:54
|
show 3 more comments
$begingroup$
We can't even do induction on a total order if it's not well-ordered. Like on $Bbb Q$ or $Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
answered Mar 20 at 8:56
ArthurArthur
121k7121207
$endgroup$
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
$begingroup$
@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
$endgroup$
– Arthur
Mar 20 at 13:54
|
show 3 more comments
$begingroup$
We can't even do induction on a total order if it's not well-ordered. Like on $Bbb Q$ or $Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
answered Mar 20 at 8:56
ArthurArthur
121k7121207
$endgroup$
We can't even do induction on a total order if it's not well-ordered. Like on $Bbb Q$ or $Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
answered Mar 20 at 8:56
ArthurArthur
121k7121207
edited Mar 20 at 13:52
answered Mar 20 at 8:56
ArthurArthur
121k7121207
answered Mar 20 at 8:56
ArthurArthur
121k7121207
answered Mar 20 at 8:56
ArthurArthur
121k7121207
121k7121207
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
$begingroup$
@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
$endgroup$
– Arthur
Mar 20 at 13:54
|
show 3 more comments
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
$begingroup$
@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
$endgroup$
– Arthur
Mar 20 at 13:54
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
$begingroup$
It is nice. I am talking that relation $rho$ which makes $X$ poset but well-ordered. Can we have induction on $X$ with respect to $rho$ ?
$endgroup$
– M. A. SARKAR
Mar 20 at 9:08
1
1
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@Arthur You can do some inductions in some totally ordered sets that are not well-ordered. For example, on the reals. Therefore, your first and second sentences are false. The third sentence is also false, since you can also do induction in well-founded sets, which might not be totaly ordered.
$endgroup$
– user647486
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
$begingroup$
@user, my question is for arbitrary poset
$endgroup$
– M. A. SARKAR
Mar 20 at 9:15
1
1
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
$endgroup$
– Peter LeFanu Lumsdaine
Mar 20 at 13:28
$begingroup$
@user647486: you can do induction on any set, if you are allowed to choose your own definition of “doing induction”. The “induction on real numbers” you link is a very nice principle, but it isn’t the most standard sense of “doing induction”, which (as this answer correctly says) fails over the reals and other non-well-founded orderings. The only shortcoming in this answer is its suggestion that well-foundedness and induction for partial orders are something novel or speculative — in fact they’re long-established and well-studied.
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– Peter LeFanu Lumsdaine
Mar 20 at 13:28
1
1
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@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
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– Arthur
Mar 20 at 13:54
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@user647486 "no difference between the statement of induction on the reals and strong induction" seems a bit much. I agree that one can do induction-like arguments on the reals under certain conditions, but I think it's a bit of a stretch to say it's the same as conventional induction.
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– Arthur
Mar 20 at 13:54
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show 3 more comments
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Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m in S$ such that there is no $x in S$ with $x R m$.
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set.
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
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So on totally ordered set , we can do induction
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– M. A. SARKAR
Mar 20 at 9:17
3
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No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
add a comment |
$begingroup$
Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m in S$ such that there is no $x in S$ with $x R m$.
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set.
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
$endgroup$
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
3
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
add a comment |
$begingroup$
Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m in S$ such that there is no $x in S$ with $x R m$.
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set.
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
$endgroup$
Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m in S$ such that there is no $x in S$ with $x R m$.
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set.
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
edited Mar 21 at 8:45
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
answered Mar 20 at 9:04
Daniel AhlsénDaniel Ahlsén
3965
3965
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
3
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
add a comment |
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
3
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
$begingroup$
So on totally ordered set , we can do induction
$endgroup$
– M. A. SARKAR
Mar 20 at 9:17
3
3
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
$begingroup$
No, not in general, since not all total orders are well-founded. The standard order on the integers is one example. Well-foundedness is equivalent to well-orderedness for total orders.
$endgroup$
– Daniel Ahlsén
Mar 20 at 10:22
add a comment |
$begingroup$
One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $xnotin S$ it is true that $forall x: (forall y: (y<xto P(y))to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
$endgroup$
add a comment |
$begingroup$
One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $xnotin S$ it is true that $forall x: (forall y: (y<xto P(y))to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
$endgroup$
add a comment |
$begingroup$
One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $xnotin S$ it is true that $forall x: (forall y: (y<xto P(y))to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
$endgroup$
One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $xnotin S$ it is true that $forall x: (forall y: (y<xto P(y))to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
answered Mar 21 at 7:58
Marc van LeeuwenMarc van Leeuwen
88.6k5111229
88.6k5111229
add a comment |
add a comment |
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$begingroup$
Supposing the answer were "yes", why do you think one would have stated the principle for well-orders in the first place? Definitions and hypotheses usually have a reason.
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:56
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@MarcvanLeeuwen, I did not get you. Can you elaborate it?
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– M. A. SARKAR
Mar 21 at 4:26
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For comparison, if the law says that you may drive car provided you have a valid driver's licence, and you ask whether more generally people can legally drive when they have no driver's licence, then you can expect a negative answer. If the answer were "yes", what would be the point of a driver's licence in the first place ?
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– Marc van Leeuwen
Mar 21 at 4:57
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@MarcvanLeeuwen, Can you elaborate it mathematically just like other users rather than your philosophically view?
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– M. A. SARKAR
Mar 21 at 6:44
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Mine was not a technical mathematical comment, so no I cannot elaborate on it mathematically. If you don't like my comments, feel free to ignore them.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:41