Mixing
A net vs a sequence
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A net is defined as a map $Thetato mathbbX$ ($thetamapsto x_theta$) where $Theta$ is a directed set and $mathbbX$ is some topological space. If $Theta=mathbbN$ then this definition coincides with the usual definition of a sequence. What would be an example of a net which is not a sequence? In particular for sequences we have that two different indices say $kneq j$ map to the same element $xin mathbbX$ i.e. $x_k=x_j$. But the same index cannot be mapped to two different elements in $mathbbX$ i.e. we cannot have $x_k=y_k$ where $x$ and $y$ are different. Is this the criteria which differentiates sequences from nets in general? Or do I get all this wrong?
real-analysis general-topology nets
asked Aug 6 at 19:47
Arian
5,235817
add a comment |Â
up vote
0
down vote
favorite
A net is defined as a map $Thetato mathbbX$ ($thetamapsto x_theta$) where $Theta$ is a directed set and $mathbbX$ is some topological space. If $Theta=mathbbN$ then this definition coincides with the usual definition of a sequence. What would be an example of a net which is not a sequence? In particular for sequences we have that two different indices say $kneq j$ map to the same element $xin mathbbX$ i.e. $x_k=x_j$. But the same index cannot be mapped to two different elements in $mathbbX$ i.e. we cannot have $x_k=y_k$ where $x$ and $y$ are different. Is this the criteria which differentiates sequences from nets in general? Or do I get all this wrong?
real-analysis general-topology nets
asked Aug 6 at 19:47
Arian
5,235817
The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A net is defined as a map $Thetato mathbbX$ ($thetamapsto x_theta$) where $Theta$ is a directed set and $mathbbX$ is some topological space. If $Theta=mathbbN$ then this definition coincides with the usual definition of a sequence. What would be an example of a net which is not a sequence? In particular for sequences we have that two different indices say $kneq j$ map to the same element $xin mathbbX$ i.e. $x_k=x_j$. But the same index cannot be mapped to two different elements in $mathbbX$ i.e. we cannot have $x_k=y_k$ where $x$ and $y$ are different. Is this the criteria which differentiates sequences from nets in general? Or do I get all this wrong?
real-analysis general-topology nets
asked Aug 6 at 19:47
Arian
5,235817
A net is defined as a map $Thetato mathbbX$ ($thetamapsto x_theta$) where $Theta$ is a directed set and $mathbbX$ is some topological space. If $Theta=mathbbN$ then this definition coincides with the usual definition of a sequence. What would be an example of a net which is not a sequence? In particular for sequences we have that two different indices say $kneq j$ map to the same element $xin mathbbX$ i.e. $x_k=x_j$. But the same index cannot be mapped to two different elements in $mathbbX$ i.e. we cannot have $x_k=y_k$ where $x$ and $y$ are different. Is this the criteria which differentiates sequences from nets in general? Or do I get all this wrong?
real-analysis general-topology nets
asked Aug 6 at 19:47
Arian
5,235817
asked Aug 6 at 19:47
Arian
5,235817
asked Aug 6 at 19:47
Arian
5,235817
asked Aug 6 at 19:47
Arian
5,235817
5,235817
The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33
add a comment |Â
The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33
The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Question: What would be an example of a net which is not a sequence?
Answer: The identity map $id:[0,1] to [0,1]$ is a net that is not a sequence.
EDIT:
Definitions:
A net is a function from a directed set $J$ to a topological space $X.$
A sequence is a function from $mathbbN$ to a topological space $X.$
Consequently, any net where $J$ is not $mathbbN$ is not a sequence.
answered Aug 6 at 20:28
Pawel
2,924921
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
add a comment |Â
up vote
4
down vote
An example that's important in proving that nets do the good things that they do: Say $X$ is a topological space and $pin X$. Let $Theta$ be the collection of all neighborhoods of $p$, ordered by reverse inclusion (so $Vge W$ if $Vsubset W$.
Recall that if $(x_alpha)$ is a net in $X$ we say that $x_alphato x$ if for every neighborhood $V$ of $x$ there exists $beta$ such that $x_alphain V$ for all $alphagebeta$. One of the reasons nets are useful is this:
Theorem Suppose that $X$ and $Y$ are topological spaces and $f:Xto Y$. Then $f$ is continuous at $xin X$ if and only if $f(x_alpha)to f(x)$ for every net $(x_alpha)subset X$ with $x_alphato x$.
If $f$ is continuous at $x$ and $x_alphato x$ then it's trivial from the definitions that $f(x_alpha)to f(x)$. For the converse we need nets defined using that funny ordered set above.
Say $f$ is not continuous at $x$. So there exists $Usubset Y$ open with $f(x)in U$ such that $f^-1(U)$ does not contain a neighborhood of $x$. Say $Theta$ is the set of all neighborhoods of $x$, ordered by reverse inclusion. For every $VinTheta$ there exists $x_Vin V$ with $f(x_V)notin U$. So $(x_V)$ is a net in $X$, and it's easy to verify that $x_Vto x$ but $f(x_V)notto f(x)$.
(The reason $Theta$ was ordered by reverse inclusion was so we could show that $x_Vto x$: Say $W$ is a neighborhood of $x$. If $Vge W$ then $x_Vin Vsubset W$, hence $x_Vin W$ for every $Vge W$.)
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
add a comment |Â
up vote
2
down vote
You already have two examples, but here's another that comes up in calculus (but usually isn't mentioned there). Let $[a,b]$ be some interval of real numbers, and let $J$ be the set of all partitions of $[a,b]$ into finitely many subintervals. Then $J$ is directed set with respect to refinement, i.e., we say that one partition is $geq$ another if the former is a refinement of the latter. Given any bounded function $f:[a,b]tomathbb R$, we can associate to each partition $jin J$ the upper Darboux sum of $f$ with respect to this partition $j$. This function, from $J$ into $mathbb R$, is a net. If it converges, then its limit deserves to be called the upper Riemann integral of $f$ over $[a,b]$. (There's another net, with the same directed set, using lower Darboux sums; its limit would be the lower Riemann integral.)
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Question: What would be an example of a net which is not a sequence?
Answer: The identity map $id:[0,1] to [0,1]$ is a net that is not a sequence.
EDIT:
Definitions:
A net is a function from a directed set $J$ to a topological space $X.$
A sequence is a function from $mathbbN$ to a topological space $X.$
Consequently, any net where $J$ is not $mathbbN$ is not a sequence.
answered Aug 6 at 20:28
Pawel
2,924921
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
add a comment |Â
up vote
0
down vote
accepted
Question: What would be an example of a net which is not a sequence?
Answer: The identity map $id:[0,1] to [0,1]$ is a net that is not a sequence.
EDIT:
Definitions:
A net is a function from a directed set $J$ to a topological space $X.$
A sequence is a function from $mathbbN$ to a topological space $X.$
Consequently, any net where $J$ is not $mathbbN$ is not a sequence.
answered Aug 6 at 20:28
Pawel
2,924921
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Question: What would be an example of a net which is not a sequence?
Answer: The identity map $id:[0,1] to [0,1]$ is a net that is not a sequence.
EDIT:
Definitions:
A net is a function from a directed set $J$ to a topological space $X.$
A sequence is a function from $mathbbN$ to a topological space $X.$
Consequently, any net where $J$ is not $mathbbN$ is not a sequence.
answered Aug 6 at 20:28
Pawel
2,924921
Question: What would be an example of a net which is not a sequence?
Answer: The identity map $id:[0,1] to [0,1]$ is a net that is not a sequence.
EDIT:
Definitions:
A net is a function from a directed set $J$ to a topological space $X.$
A sequence is a function from $mathbbN$ to a topological space $X.$
Consequently, any net where $J$ is not $mathbbN$ is not a sequence.
answered Aug 6 at 20:28
Pawel
2,924921
edited Aug 6 at 21:19
answered Aug 6 at 20:28
Pawel
2,924921
answered Aug 6 at 20:28
Pawel
2,924921
answered Aug 6 at 20:28
Pawel
2,924921
2,924921
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
add a comment |Â
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
Ok I see that here $(x_r)$ with $r$ from the interval $[0,1]$. Like $x_0.234$ for example. Did I get this right? If this is correct why wouldn't it be a sequence? Is it because there is no bijective map between $[0,1]$ and $mathbbN$?
– Arian
Aug 6 at 20:32
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
I tried to edit my answer to make everything clearer.
– Pawel
Aug 6 at 20:38
Thanks @Pawel..
– Arian
Aug 6 at 20:39
Thanks @Pawel..
– Arian
Aug 6 at 20:39
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
@DavidC.Ullrich You are correct. I will edit my answer. Thank you for pointing it out.
– Pawel
Aug 6 at 21:18
add a comment |Â
up vote
4
down vote
An example that's important in proving that nets do the good things that they do: Say $X$ is a topological space and $pin X$. Let $Theta$ be the collection of all neighborhoods of $p$, ordered by reverse inclusion (so $Vge W$ if $Vsubset W$.
Recall that if $(x_alpha)$ is a net in $X$ we say that $x_alphato x$ if for every neighborhood $V$ of $x$ there exists $beta$ such that $x_alphain V$ for all $alphagebeta$. One of the reasons nets are useful is this:
Theorem Suppose that $X$ and $Y$ are topological spaces and $f:Xto Y$. Then $f$ is continuous at $xin X$ if and only if $f(x_alpha)to f(x)$ for every net $(x_alpha)subset X$ with $x_alphato x$.
If $f$ is continuous at $x$ and $x_alphato x$ then it's trivial from the definitions that $f(x_alpha)to f(x)$. For the converse we need nets defined using that funny ordered set above.
Say $f$ is not continuous at $x$. So there exists $Usubset Y$ open with $f(x)in U$ such that $f^-1(U)$ does not contain a neighborhood of $x$. Say $Theta$ is the set of all neighborhoods of $x$, ordered by reverse inclusion. For every $VinTheta$ there exists $x_Vin V$ with $f(x_V)notin U$. So $(x_V)$ is a net in $X$, and it's easy to verify that $x_Vto x$ but $f(x_V)notto f(x)$.
(The reason $Theta$ was ordered by reverse inclusion was so we could show that $x_Vto x$: Say $W$ is a neighborhood of $x$. If $Vge W$ then $x_Vin Vsubset W$, hence $x_Vin W$ for every $Vge W$.)
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
add a comment |Â
up vote
4
down vote
An example that's important in proving that nets do the good things that they do: Say $X$ is a topological space and $pin X$. Let $Theta$ be the collection of all neighborhoods of $p$, ordered by reverse inclusion (so $Vge W$ if $Vsubset W$.
Recall that if $(x_alpha)$ is a net in $X$ we say that $x_alphato x$ if for every neighborhood $V$ of $x$ there exists $beta$ such that $x_alphain V$ for all $alphagebeta$. One of the reasons nets are useful is this:
Theorem Suppose that $X$ and $Y$ are topological spaces and $f:Xto Y$. Then $f$ is continuous at $xin X$ if and only if $f(x_alpha)to f(x)$ for every net $(x_alpha)subset X$ with $x_alphato x$.
If $f$ is continuous at $x$ and $x_alphato x$ then it's trivial from the definitions that $f(x_alpha)to f(x)$. For the converse we need nets defined using that funny ordered set above.
Say $f$ is not continuous at $x$. So there exists $Usubset Y$ open with $f(x)in U$ such that $f^-1(U)$ does not contain a neighborhood of $x$. Say $Theta$ is the set of all neighborhoods of $x$, ordered by reverse inclusion. For every $VinTheta$ there exists $x_Vin V$ with $f(x_V)notin U$. So $(x_V)$ is a net in $X$, and it's easy to verify that $x_Vto x$ but $f(x_V)notto f(x)$.
(The reason $Theta$ was ordered by reverse inclusion was so we could show that $x_Vto x$: Say $W$ is a neighborhood of $x$. If $Vge W$ then $x_Vin Vsubset W$, hence $x_Vin W$ for every $Vge W$.)
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
add a comment |Â
up vote
4
down vote
up vote
4
down vote
An example that's important in proving that nets do the good things that they do: Say $X$ is a topological space and $pin X$. Let $Theta$ be the collection of all neighborhoods of $p$, ordered by reverse inclusion (so $Vge W$ if $Vsubset W$.
Recall that if $(x_alpha)$ is a net in $X$ we say that $x_alphato x$ if for every neighborhood $V$ of $x$ there exists $beta$ such that $x_alphain V$ for all $alphagebeta$. One of the reasons nets are useful is this:
Theorem Suppose that $X$ and $Y$ are topological spaces and $f:Xto Y$. Then $f$ is continuous at $xin X$ if and only if $f(x_alpha)to f(x)$ for every net $(x_alpha)subset X$ with $x_alphato x$.
If $f$ is continuous at $x$ and $x_alphato x$ then it's trivial from the definitions that $f(x_alpha)to f(x)$. For the converse we need nets defined using that funny ordered set above.
Say $f$ is not continuous at $x$. So there exists $Usubset Y$ open with $f(x)in U$ such that $f^-1(U)$ does not contain a neighborhood of $x$. Say $Theta$ is the set of all neighborhoods of $x$, ordered by reverse inclusion. For every $VinTheta$ there exists $x_Vin V$ with $f(x_V)notin U$. So $(x_V)$ is a net in $X$, and it's easy to verify that $x_Vto x$ but $f(x_V)notto f(x)$.
(The reason $Theta$ was ordered by reverse inclusion was so we could show that $x_Vto x$: Say $W$ is a neighborhood of $x$. If $Vge W$ then $x_Vin Vsubset W$, hence $x_Vin W$ for every $Vge W$.)
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
An example that's important in proving that nets do the good things that they do: Say $X$ is a topological space and $pin X$. Let $Theta$ be the collection of all neighborhoods of $p$, ordered by reverse inclusion (so $Vge W$ if $Vsubset W$.
Recall that if $(x_alpha)$ is a net in $X$ we say that $x_alphato x$ if for every neighborhood $V$ of $x$ there exists $beta$ such that $x_alphain V$ for all $alphagebeta$. One of the reasons nets are useful is this:
Theorem Suppose that $X$ and $Y$ are topological spaces and $f:Xto Y$. Then $f$ is continuous at $xin X$ if and only if $f(x_alpha)to f(x)$ for every net $(x_alpha)subset X$ with $x_alphato x$.
If $f$ is continuous at $x$ and $x_alphato x$ then it's trivial from the definitions that $f(x_alpha)to f(x)$. For the converse we need nets defined using that funny ordered set above.
Say $f$ is not continuous at $x$. So there exists $Usubset Y$ open with $f(x)in U$ such that $f^-1(U)$ does not contain a neighborhood of $x$. Say $Theta$ is the set of all neighborhoods of $x$, ordered by reverse inclusion. For every $VinTheta$ there exists $x_Vin V$ with $f(x_V)notin U$. So $(x_V)$ is a net in $X$, and it's easy to verify that $x_Vto x$ but $f(x_V)notto f(x)$.
(The reason $Theta$ was ordered by reverse inclusion was so we could show that $x_Vto x$: Say $W$ is a neighborhood of $x$. If $Vge W$ then $x_Vin Vsubset W$, hence $x_Vin W$ for every $Vge W$.)
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
edited Aug 7 at 15:54
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
answered Aug 6 at 20:21
David C. Ullrich
54.4k33684
54.4k33684
add a comment |Â
add a comment |Â
up vote
2
down vote
You already have two examples, but here's another that comes up in calculus (but usually isn't mentioned there). Let $[a,b]$ be some interval of real numbers, and let $J$ be the set of all partitions of $[a,b]$ into finitely many subintervals. Then $J$ is directed set with respect to refinement, i.e., we say that one partition is $geq$ another if the former is a refinement of the latter. Given any bounded function $f:[a,b]tomathbb R$, we can associate to each partition $jin J$ the upper Darboux sum of $f$ with respect to this partition $j$. This function, from $J$ into $mathbb R$, is a net. If it converges, then its limit deserves to be called the upper Riemann integral of $f$ over $[a,b]$. (There's another net, with the same directed set, using lower Darboux sums; its limit would be the lower Riemann integral.)
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
add a comment |Â
up vote
2
down vote
You already have two examples, but here's another that comes up in calculus (but usually isn't mentioned there). Let $[a,b]$ be some interval of real numbers, and let $J$ be the set of all partitions of $[a,b]$ into finitely many subintervals. Then $J$ is directed set with respect to refinement, i.e., we say that one partition is $geq$ another if the former is a refinement of the latter. Given any bounded function $f:[a,b]tomathbb R$, we can associate to each partition $jin J$ the upper Darboux sum of $f$ with respect to this partition $j$. This function, from $J$ into $mathbb R$, is a net. If it converges, then its limit deserves to be called the upper Riemann integral of $f$ over $[a,b]$. (There's another net, with the same directed set, using lower Darboux sums; its limit would be the lower Riemann integral.)
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You already have two examples, but here's another that comes up in calculus (but usually isn't mentioned there). Let $[a,b]$ be some interval of real numbers, and let $J$ be the set of all partitions of $[a,b]$ into finitely many subintervals. Then $J$ is directed set with respect to refinement, i.e., we say that one partition is $geq$ another if the former is a refinement of the latter. Given any bounded function $f:[a,b]tomathbb R$, we can associate to each partition $jin J$ the upper Darboux sum of $f$ with respect to this partition $j$. This function, from $J$ into $mathbb R$, is a net. If it converges, then its limit deserves to be called the upper Riemann integral of $f$ over $[a,b]$. (There's another net, with the same directed set, using lower Darboux sums; its limit would be the lower Riemann integral.)
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
You already have two examples, but here's another that comes up in calculus (but usually isn't mentioned there). Let $[a,b]$ be some interval of real numbers, and let $J$ be the set of all partitions of $[a,b]$ into finitely many subintervals. Then $J$ is directed set with respect to refinement, i.e., we say that one partition is $geq$ another if the former is a refinement of the latter. Given any bounded function $f:[a,b]tomathbb R$, we can associate to each partition $jin J$ the upper Darboux sum of $f$ with respect to this partition $j$. This function, from $J$ into $mathbb R$, is a net. If it converges, then its limit deserves to be called the upper Riemann integral of $f$ over $[a,b]$. (There's another net, with the same directed set, using lower Darboux sums; its limit would be the lower Riemann integral.)
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
answered Aug 6 at 22:24
Andreas Blass
47.6k348104
47.6k348104
add a comment |Â
add a comment |Â
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The fact that it's a mapping, ie a function, says that the same index cannot be mapped to two different elements.
– David C. Ullrich
Aug 6 at 20:26
Thanks @DavidC.Ullrich for this clarification.
– Arian
Aug 6 at 20:33