A net has a limit if and only if all of its subnets have limits (without the use of Cauchy nets)
$begingroup$
I am trying to prove the above. More specifically I am trying to prove the direction "if all subnets of a given net have limits then the net in question has a limit"
The definition I am using for a subnet is as follows (From Folland)
A subnet of a net $(x_alpha)_{alphain A}$ is a net $(y_{beta})_{betain B}$ together with a map $h:Brightarrow A$ such that
-For every $alpha_{0}in A$, there exists $beta_{0}in B$ such that $forall beta$ such that $beta_{0}preccurlyeq beta$ we have $alpha_{0}preccurlyeq h(beta)$
-$y_{beta}=x_{h(alpha)}$
I feel the only way I can approach this is via contradiction. Suppose that $(x_alpha)_{alphain A}$ did not have a limit, then...then what. I feel I need a way to talk about convergence without knowing what the limit is. In real analysis, one could do this using Cauchy sequences. A quick look on wikipedia shows that a thing called a Cauchy net exists, but since my lecturer gave us this problem without ever talking about Cauchy nets I feel there should be a way to solve this without referring to Cauchy nets.
To be clear, the only things we were taught was basic point set topology and the definition of a net and a subnet, so I would like, if possible, an answer that only utilises these things. A hint or complete answer is welcome.
general-topology nets
asked Dec 11 '18 at 1:39
DamoDamo
486210
$endgroup$
add a comment |
$begingroup$
I am trying to prove the above. More specifically I am trying to prove the direction "if all subnets of a given net have limits then the net in question has a limit"
The definition I am using for a subnet is as follows (From Folland)
A subnet of a net $(x_alpha)_{alphain A}$ is a net $(y_{beta})_{betain B}$ together with a map $h:Brightarrow A$ such that
-For every $alpha_{0}in A$, there exists $beta_{0}in B$ such that $forall beta$ such that $beta_{0}preccurlyeq beta$ we have $alpha_{0}preccurlyeq h(beta)$
-$y_{beta}=x_{h(alpha)}$
I feel the only way I can approach this is via contradiction. Suppose that $(x_alpha)_{alphain A}$ did not have a limit, then...then what. I feel I need a way to talk about convergence without knowing what the limit is. In real analysis, one could do this using Cauchy sequences. A quick look on wikipedia shows that a thing called a Cauchy net exists, but since my lecturer gave us this problem without ever talking about Cauchy nets I feel there should be a way to solve this without referring to Cauchy nets.
To be clear, the only things we were taught was basic point set topology and the definition of a net and a subnet, so I would like, if possible, an answer that only utilises these things. A hint or complete answer is welcome.
general-topology nets
asked Dec 11 '18 at 1:39
DamoDamo
486210
$endgroup$
2
$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
1
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00
add a comment |
$begingroup$
I am trying to prove the above. More specifically I am trying to prove the direction "if all subnets of a given net have limits then the net in question has a limit"
The definition I am using for a subnet is as follows (From Folland)
A subnet of a net $(x_alpha)_{alphain A}$ is a net $(y_{beta})_{betain B}$ together with a map $h:Brightarrow A$ such that
-For every $alpha_{0}in A$, there exists $beta_{0}in B$ such that $forall beta$ such that $beta_{0}preccurlyeq beta$ we have $alpha_{0}preccurlyeq h(beta)$
-$y_{beta}=x_{h(alpha)}$
I feel the only way I can approach this is via contradiction. Suppose that $(x_alpha)_{alphain A}$ did not have a limit, then...then what. I feel I need a way to talk about convergence without knowing what the limit is. In real analysis, one could do this using Cauchy sequences. A quick look on wikipedia shows that a thing called a Cauchy net exists, but since my lecturer gave us this problem without ever talking about Cauchy nets I feel there should be a way to solve this without referring to Cauchy nets.
To be clear, the only things we were taught was basic point set topology and the definition of a net and a subnet, so I would like, if possible, an answer that only utilises these things. A hint or complete answer is welcome.
general-topology nets
asked Dec 11 '18 at 1:39
DamoDamo
486210
$endgroup$
I am trying to prove the above. More specifically I am trying to prove the direction "if all subnets of a given net have limits then the net in question has a limit"
The definition I am using for a subnet is as follows (From Folland)
A subnet of a net $(x_alpha)_{alphain A}$ is a net $(y_{beta})_{betain B}$ together with a map $h:Brightarrow A$ such that
-For every $alpha_{0}in A$, there exists $beta_{0}in B$ such that $forall beta$ such that $beta_{0}preccurlyeq beta$ we have $alpha_{0}preccurlyeq h(beta)$
-$y_{beta}=x_{h(alpha)}$
I feel the only way I can approach this is via contradiction. Suppose that $(x_alpha)_{alphain A}$ did not have a limit, then...then what. I feel I need a way to talk about convergence without knowing what the limit is. In real analysis, one could do this using Cauchy sequences. A quick look on wikipedia shows that a thing called a Cauchy net exists, but since my lecturer gave us this problem without ever talking about Cauchy nets I feel there should be a way to solve this without referring to Cauchy nets.
To be clear, the only things we were taught was basic point set topology and the definition of a net and a subnet, so I would like, if possible, an answer that only utilises these things. A hint or complete answer is welcome.
general-topology nets
general-topology nets
asked Dec 11 '18 at 1:39
DamoDamo
486210
asked Dec 11 '18 at 1:39
DamoDamo
486210
asked Dec 11 '18 at 1:39
DamoDamo
486210
asked Dec 11 '18 at 1:39
DamoDamo
486210
asked Dec 11 '18 at 1:39
DamoDamo
486210
486210
2
$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
1
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00
add a comment |
2
$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
1
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00
2
2
$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
1
1
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00
add a comment |
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$begingroup$
What about the fact that any net $(x_alpha)_{alpha in A}$ is a subnet of itself, via the identity map $A to A$?
$endgroup$
– Daniel Schepler
Dec 11 '18 at 1:45
$begingroup$
My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me.
$endgroup$
– Damo
Dec 11 '18 at 1:49
1
$begingroup$
I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence.
$endgroup$
– Song
Dec 11 '18 at 2:00