Mixing
Question about the completeness axiom
$begingroup$
I have been looking online and on lecure notes and I have observed that there are 2 definitions for the completeness axiom and I cannot relate them together.
These are:
1) Every non-empty set of real numbers that is bounded above has
a supremum. Every non-empty set of real numbers that is bounded
below has an infimum.
2) Every real sequence that is increasing and bounded above will converge.
What I am thinking about is that if there exists a supremum/infimum then every sequence will have to converge specifically to that supremum/infimum.
Would that be the idea of how 1 relates to 2?
supremum-and-infimum
asked Jan 8 at 0:08
ValVal
557
$endgroup$
add a comment |
$begingroup$
I have been looking online and on lecure notes and I have observed that there are 2 definitions for the completeness axiom and I cannot relate them together.
These are:
1) Every non-empty set of real numbers that is bounded above has
a supremum. Every non-empty set of real numbers that is bounded
below has an infimum.
2) Every real sequence that is increasing and bounded above will converge.
What I am thinking about is that if there exists a supremum/infimum then every sequence will have to converge specifically to that supremum/infimum.
Would that be the idea of how 1 relates to 2?
supremum-and-infimum
asked Jan 8 at 0:08
ValVal
557
$endgroup$
$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
$endgroup$
– Matt Samuel
Jan 8 at 0:13
$begingroup$
Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
$endgroup$
– Val
Jan 8 at 0:18
$begingroup$
The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
$endgroup$
– Matt Samuel
Jan 8 at 0:21
$begingroup$
You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
$endgroup$
– timtfj
Jan 8 at 0:31
add a comment |
$begingroup$
I have been looking online and on lecure notes and I have observed that there are 2 definitions for the completeness axiom and I cannot relate them together.
These are:
1) Every non-empty set of real numbers that is bounded above has
a supremum. Every non-empty set of real numbers that is bounded
below has an infimum.
2) Every real sequence that is increasing and bounded above will converge.
What I am thinking about is that if there exists a supremum/infimum then every sequence will have to converge specifically to that supremum/infimum.
Would that be the idea of how 1 relates to 2?
supremum-and-infimum
asked Jan 8 at 0:08
ValVal
557
$endgroup$
I have been looking online and on lecure notes and I have observed that there are 2 definitions for the completeness axiom and I cannot relate them together.
These are:
1) Every non-empty set of real numbers that is bounded above has
a supremum. Every non-empty set of real numbers that is bounded
below has an infimum.
2) Every real sequence that is increasing and bounded above will converge.
What I am thinking about is that if there exists a supremum/infimum then every sequence will have to converge specifically to that supremum/infimum.
Would that be the idea of how 1 relates to 2?
supremum-and-infimum
supremum-and-infimum
asked Jan 8 at 0:08
ValVal
557
asked Jan 8 at 0:08
ValVal
557
edited Jan 8 at 0:13
Val
asked Jan 8 at 0:08
ValVal
557
asked Jan 8 at 0:08
ValVal
557
asked Jan 8 at 0:08
ValVal
557
557
$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
$endgroup$
– Matt Samuel
Jan 8 at 0:13
$begingroup$
Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
$endgroup$
– Val
Jan 8 at 0:18
$begingroup$
The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
$endgroup$
– Matt Samuel
Jan 8 at 0:21
$begingroup$
You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
$endgroup$
– timtfj
Jan 8 at 0:31
add a comment |
$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
$endgroup$
– Matt Samuel
Jan 8 at 0:13
$begingroup$
Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
$endgroup$
– Val
Jan 8 at 0:18
$begingroup$
The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
$endgroup$
– Matt Samuel
Jan 8 at 0:21
$begingroup$
You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
$endgroup$
– timtfj
Jan 8 at 0:31
$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
$endgroup$
– Matt Samuel
Jan 8 at 0:13
$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
$endgroup$
– Matt Samuel
Jan 8 at 0:13
$begingroup$
Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
$endgroup$
– Val
Jan 8 at 0:18
$begingroup$
Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
$endgroup$
– Val
Jan 8 at 0:18
$begingroup$
The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
$endgroup$
– Matt Samuel
Jan 8 at 0:21
$begingroup$
The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
$endgroup$
– Matt Samuel
Jan 8 at 0:21
$begingroup$
You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
$endgroup$
– timtfj
Jan 8 at 0:31
$begingroup$
You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
$endgroup$
– timtfj
Jan 8 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $ge$ every $x_n$. Since it's the least upper bound, $L-epsilon$ is not an upper bound (for an arbitrary $epsilon>0$) - which means that there is some $N$ such that $x_N > L-epsilon$. Then, for all $n>N$, $L-epsilon < x_N le x_nle L$ and $|x_n-L| <epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $ale b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=frac{x_0+y_0}{2}$. If no, let $x_1=frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=frac{x_n+y_n}{2}$ if $frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=lim_n x_n$ and $y=lim_n y_n$. In addition, $y_n-x_n=frac{1}{2^n}(b-a)to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $cin S$ is $le$ each $y_n$, and thus also $le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
answered Jan 8 at 3:23
jmerryjmerry
5,857617
$endgroup$
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
$endgroup$
– timtfj
Jan 19 at 15:56
add a comment |
$begingroup$
If an increasing sequence has a supremum, then it is bounded above and by (2) it does converge to the supremum.
Similarly if a decreasing sequence has an infimum, then it is bounded below and it converges to its infimum.
Thus your statement is valid for monotonic sequences that the existence of supremum or infimum implies the convergence to the supremum or infimum.
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
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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
$begingroup$
There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $ge$ every $x_n$. Since it's the least upper bound, $L-epsilon$ is not an upper bound (for an arbitrary $epsilon>0$) - which means that there is some $N$ such that $x_N > L-epsilon$. Then, for all $n>N$, $L-epsilon < x_N le x_nle L$ and $|x_n-L| <epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $ale b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=frac{x_0+y_0}{2}$. If no, let $x_1=frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=frac{x_n+y_n}{2}$ if $frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=lim_n x_n$ and $y=lim_n y_n$. In addition, $y_n-x_n=frac{1}{2^n}(b-a)to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $cin S$ is $le$ each $y_n$, and thus also $le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
answered Jan 8 at 3:23
jmerryjmerry
5,857617
$endgroup$
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
$endgroup$
– timtfj
Jan 19 at 15:56
add a comment |
$begingroup$
There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $ge$ every $x_n$. Since it's the least upper bound, $L-epsilon$ is not an upper bound (for an arbitrary $epsilon>0$) - which means that there is some $N$ such that $x_N > L-epsilon$. Then, for all $n>N$, $L-epsilon < x_N le x_nle L$ and $|x_n-L| <epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $ale b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=frac{x_0+y_0}{2}$. If no, let $x_1=frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=frac{x_n+y_n}{2}$ if $frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=lim_n x_n$ and $y=lim_n y_n$. In addition, $y_n-x_n=frac{1}{2^n}(b-a)to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $cin S$ is $le$ each $y_n$, and thus also $le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
answered Jan 8 at 3:23
jmerryjmerry
5,857617
$endgroup$
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
$endgroup$
– timtfj
Jan 19 at 15:56
add a comment |
$begingroup$
There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $ge$ every $x_n$. Since it's the least upper bound, $L-epsilon$ is not an upper bound (for an arbitrary $epsilon>0$) - which means that there is some $N$ such that $x_N > L-epsilon$. Then, for all $n>N$, $L-epsilon < x_N le x_nle L$ and $|x_n-L| <epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $ale b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=frac{x_0+y_0}{2}$. If no, let $x_1=frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=frac{x_n+y_n}{2}$ if $frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=lim_n x_n$ and $y=lim_n y_n$. In addition, $y_n-x_n=frac{1}{2^n}(b-a)to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $cin S$ is $le$ each $y_n$, and thus also $le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
answered Jan 8 at 3:23
jmerryjmerry
5,857617
$endgroup$
There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $ge$ every $x_n$. Since it's the least upper bound, $L-epsilon$ is not an upper bound (for an arbitrary $epsilon>0$) - which means that there is some $N$ such that $x_N > L-epsilon$. Then, for all $n>N$, $L-epsilon < x_N le x_nle L$ and $|x_n-L| <epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $ale b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=frac{x_0+y_0}{2}$. If no, let $x_1=frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=frac{x_n+y_n}{2}$ if $frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=lim_n x_n$ and $y=lim_n y_n$. In addition, $y_n-x_n=frac{1}{2^n}(b-a)to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $cin S$ is $le$ each $y_n$, and thus also $le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
answered Jan 8 at 3:23
jmerryjmerry
5,857617
answered Jan 8 at 3:23
jmerryjmerry
5,857617
answered Jan 8 at 3:23
jmerryjmerry
5,857617
answered Jan 8 at 3:23
jmerryjmerry
5,857617
5,857617
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
$endgroup$
– timtfj
Jan 19 at 15:56
add a comment |
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
$endgroup$
– timtfj
Jan 19 at 15:56
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Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
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– timtfj
Jan 19 at 15:56
$begingroup$
Just adding the thought that without being aware of the equivalence of the different definitions, it's possible to find yourself stuck in a circular proof: prove $A$ using $B$, prove $B$ using $C$,, then find you need $A$ for your proof of $C$—when you should probably have just picked $A$ as the definition of completeness to use.
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– timtfj
Jan 19 at 15:56
add a comment |
$begingroup$
If an increasing sequence has a supremum, then it is bounded above and by (2) it does converge to the supremum.
Similarly if a decreasing sequence has an infimum, then it is bounded below and it converges to its infimum.
Thus your statement is valid for monotonic sequences that the existence of supremum or infimum implies the convergence to the supremum or infimum.
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
If an increasing sequence has a supremum, then it is bounded above and by (2) it does converge to the supremum.
Similarly if a decreasing sequence has an infimum, then it is bounded below and it converges to its infimum.
Thus your statement is valid for monotonic sequences that the existence of supremum or infimum implies the convergence to the supremum or infimum.
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
add a comment |
$begingroup$
If an increasing sequence has a supremum, then it is bounded above and by (2) it does converge to the supremum.
Similarly if a decreasing sequence has an infimum, then it is bounded below and it converges to its infimum.
Thus your statement is valid for monotonic sequences that the existence of supremum or infimum implies the convergence to the supremum or infimum.
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
If an increasing sequence has a supremum, then it is bounded above and by (2) it does converge to the supremum.
Similarly if a decreasing sequence has an infimum, then it is bounded below and it converges to its infimum.
Thus your statement is valid for monotonic sequences that the existence of supremum or infimum implies the convergence to the supremum or infimum.
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 8 at 2:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
add a comment |
add a comment |
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$begingroup$
That works for real numbers, but there are ordered sets where there is no sequence that converges to the supremum.
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– Matt Samuel
Jan 8 at 0:13
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Would that mean when working with real numbers 1 and 2 will always be the same definition for the fact that the supremum or infimum will be a limit?
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– Val
Jan 8 at 0:18
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The supremum of an increasing bounded sequence is its limit. That's true in general. But for example the initial section of the first uncountable ordinal has the first uncountable ordinal as its supremum, but there is no sequence that converges to the first uncountable ordinal.
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– Matt Samuel
Jan 8 at 0:21
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You can also use "Every Cauchy sequence in $mathbb R$ has a limit in $mathbb R$" and derive both versions you've given from that: first prove that the infimum/supremum exists and then that the bounded sequence converges to it. I think that's a third version of the axiom (but I may be wrong).
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– timtfj
Jan 8 at 0:31