Mixing
Question regarding proof of Tychonoff's theorem
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On Wikipedia it states that a space $X$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.
In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since ${0,1}^mathbb{R}$ is not sequentially compact, but it is compact by Tychonoff's theorem.
compactness product-space
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
$endgroup$
add a comment |
$begingroup$
On Wikipedia it states that a space $X$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.
In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since ${0,1}^mathbb{R}$ is not sequentially compact, but it is compact by Tychonoff's theorem.
compactness product-space
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
$endgroup$
1
$begingroup$
Can you give an example of a subnet that is not a subsequence?
$endgroup$
– SmileyCraft
Jan 20 at 5:40
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
$endgroup$
– SmileyCraft
Jan 20 at 5:49
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
$endgroup$
– SmileyCraft
Jan 20 at 6:03
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
$endgroup$
– David C. Ullrich
Jan 20 at 18:46
add a comment |
$begingroup$
On Wikipedia it states that a space $X$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.
In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since ${0,1}^mathbb{R}$ is not sequentially compact, but it is compact by Tychonoff's theorem.
compactness product-space
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
$endgroup$
On Wikipedia it states that a space $X$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.
In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since ${0,1}^mathbb{R}$ is not sequentially compact, but it is compact by Tychonoff's theorem.
compactness product-space
compactness product-space
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
edited Jan 20 at 5:51
SmileyCraft
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
asked Jan 20 at 5:34
SmileyCraftSmileyCraft
3,476518
3,476518
1
$begingroup$
Can you give an example of a subnet that is not a subsequence?
$endgroup$
– SmileyCraft
Jan 20 at 5:40
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
$endgroup$
– SmileyCraft
Jan 20 at 5:49
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
$endgroup$
– SmileyCraft
Jan 20 at 6:03
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
$endgroup$
– David C. Ullrich
Jan 20 at 18:46
add a comment |
1
$begingroup$
Can you give an example of a subnet that is not a subsequence?
$endgroup$
– SmileyCraft
Jan 20 at 5:40
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
$endgroup$
– SmileyCraft
Jan 20 at 5:49
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
$endgroup$
– SmileyCraft
Jan 20 at 6:03
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
$endgroup$
– David C. Ullrich
Jan 20 at 18:46
1
1
$begingroup$
Can you give an example of a subnet that is not a subsequence?
$endgroup$
– SmileyCraft
Jan 20 at 5:40
$begingroup$
Can you give an example of a subnet that is not a subsequence?
$endgroup$
– SmileyCraft
Jan 20 at 5:40
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
$endgroup$
– SmileyCraft
Jan 20 at 5:49
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
$endgroup$
– SmileyCraft
Jan 20 at 5:49
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
$endgroup$
– SmileyCraft
Jan 20 at 6:03
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
$endgroup$
– SmileyCraft
Jan 20 at 6:03
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
$endgroup$
– David C. Ullrich
Jan 20 at 18:46
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
$endgroup$
– David C. Ullrich
Jan 20 at 18:46
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.
Let $I={0,1}^mathbb N$.
The product space $X={0,1}^I$ is a compact space which is not sequentially compact.
For $ninmathbb N$ define $f_n:Ito{0,1}$ by setting $f_n(i)=i(n)$. Then $langle f_n:ninmathbb Nrangle$ is a sequence in $X$ with no convergent subsequence.
Let $mathcal U$ be a uniform ultrafilter on $mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)
Define $f:Ito{0,1}$ so that, for each $iin I$, ${ninmathbb N:i(n)=f(i)}inmathcal U$.
Let $D$ be the collection of all finite subsets of $I$, directed by $subseteq$.
For $Kin D$, let $h(K)$ be the least $ninmathbb N$ such that $i(n)=f(i)$ for all $iin K$; this defines a monotone final function $h:Dtomathbb N$.
Define $g_K=f_{h(K)}in X$; then $langle g_K:Kin Drangle$ is a subnet of $langle f_n:ninmathbb Nrangle$ which converges to $f$.
answered Jan 20 at 7:17
bofbof
52.3k558121
$endgroup$
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
add a comment |
$begingroup$
The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $omega_1$, the first uncountable ordinal, then there exists no final function $h:mathbb{N}toomega_1$, since $cup h(n)$ is an upper bound on $h(n)$.
To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence ${f_n}$ in $Pi X_{alphain A}$ as a tuple $(f,I)$ where $finPi X_{alphain A}$ and $Isubseteq A$ such that $f$ is a cluster point of ${f_n|_I}$. We can then define the partial ordening $(f,I)leq(g,J)$ iff $Isubseteq J$ and $g|_I=f$.
Then there obviously exists some partial clusterpoint; consider $(f,emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $ain Asetminus I$. Then the $a$-coordinate of the subnet ${g_n|_I}$ of ${f_n|_I}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,Icup{a})$.
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.
Let $I={0,1}^mathbb N$.
The product space $X={0,1}^I$ is a compact space which is not sequentially compact.
For $ninmathbb N$ define $f_n:Ito{0,1}$ by setting $f_n(i)=i(n)$. Then $langle f_n:ninmathbb Nrangle$ is a sequence in $X$ with no convergent subsequence.
Let $mathcal U$ be a uniform ultrafilter on $mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)
Define $f:Ito{0,1}$ so that, for each $iin I$, ${ninmathbb N:i(n)=f(i)}inmathcal U$.
Let $D$ be the collection of all finite subsets of $I$, directed by $subseteq$.
For $Kin D$, let $h(K)$ be the least $ninmathbb N$ such that $i(n)=f(i)$ for all $iin K$; this defines a monotone final function $h:Dtomathbb N$.
Define $g_K=f_{h(K)}in X$; then $langle g_K:Kin Drangle$ is a subnet of $langle f_n:ninmathbb Nrangle$ which converges to $f$.
answered Jan 20 at 7:17
bofbof
52.3k558121
$endgroup$
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
add a comment |
$begingroup$
Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.
Let $I={0,1}^mathbb N$.
The product space $X={0,1}^I$ is a compact space which is not sequentially compact.
For $ninmathbb N$ define $f_n:Ito{0,1}$ by setting $f_n(i)=i(n)$. Then $langle f_n:ninmathbb Nrangle$ is a sequence in $X$ with no convergent subsequence.
Let $mathcal U$ be a uniform ultrafilter on $mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)
Define $f:Ito{0,1}$ so that, for each $iin I$, ${ninmathbb N:i(n)=f(i)}inmathcal U$.
Let $D$ be the collection of all finite subsets of $I$, directed by $subseteq$.
For $Kin D$, let $h(K)$ be the least $ninmathbb N$ such that $i(n)=f(i)$ for all $iin K$; this defines a monotone final function $h:Dtomathbb N$.
Define $g_K=f_{h(K)}in X$; then $langle g_K:Kin Drangle$ is a subnet of $langle f_n:ninmathbb Nrangle$ which converges to $f$.
answered Jan 20 at 7:17
bofbof
52.3k558121
$endgroup$
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
add a comment |
$begingroup$
Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.
Let $I={0,1}^mathbb N$.
The product space $X={0,1}^I$ is a compact space which is not sequentially compact.
For $ninmathbb N$ define $f_n:Ito{0,1}$ by setting $f_n(i)=i(n)$. Then $langle f_n:ninmathbb Nrangle$ is a sequence in $X$ with no convergent subsequence.
Let $mathcal U$ be a uniform ultrafilter on $mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)
Define $f:Ito{0,1}$ so that, for each $iin I$, ${ninmathbb N:i(n)=f(i)}inmathcal U$.
Let $D$ be the collection of all finite subsets of $I$, directed by $subseteq$.
For $Kin D$, let $h(K)$ be the least $ninmathbb N$ such that $i(n)=f(i)$ for all $iin K$; this defines a monotone final function $h:Dtomathbb N$.
Define $g_K=f_{h(K)}in X$; then $langle g_K:Kin Drangle$ is a subnet of $langle f_n:ninmathbb Nrangle$ which converges to $f$.
answered Jan 20 at 7:17
bofbof
52.3k558121
$endgroup$
Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.
Let $I={0,1}^mathbb N$.
The product space $X={0,1}^I$ is a compact space which is not sequentially compact.
For $ninmathbb N$ define $f_n:Ito{0,1}$ by setting $f_n(i)=i(n)$. Then $langle f_n:ninmathbb Nrangle$ is a sequence in $X$ with no convergent subsequence.
Let $mathcal U$ be a uniform ultrafilter on $mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)
Define $f:Ito{0,1}$ so that, for each $iin I$, ${ninmathbb N:i(n)=f(i)}inmathcal U$.
Let $D$ be the collection of all finite subsets of $I$, directed by $subseteq$.
For $Kin D$, let $h(K)$ be the least $ninmathbb N$ such that $i(n)=f(i)$ for all $iin K$; this defines a monotone final function $h:Dtomathbb N$.
Define $g_K=f_{h(K)}in X$; then $langle g_K:Kin Drangle$ is a subnet of $langle f_n:ninmathbb Nrangle$ which converges to $f$.
answered Jan 20 at 7:17
bofbof
52.3k558121
edited Jan 20 at 9:08
answered Jan 20 at 7:17
bofbof
52.3k558121
answered Jan 20 at 7:17
bofbof
52.3k558121
answered Jan 20 at 7:17
bofbof
52.3k558121
52.3k558121
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
add a comment |
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it.
$endgroup$
– SmileyCraft
Jan 20 at 9:27
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
$begingroup$
Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful.
$endgroup$
– SmileyCraft
Jan 22 at 13:03
add a comment |
$begingroup$
The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $omega_1$, the first uncountable ordinal, then there exists no final function $h:mathbb{N}toomega_1$, since $cup h(n)$ is an upper bound on $h(n)$.
To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence ${f_n}$ in $Pi X_{alphain A}$ as a tuple $(f,I)$ where $finPi X_{alphain A}$ and $Isubseteq A$ such that $f$ is a cluster point of ${f_n|_I}$. We can then define the partial ordening $(f,I)leq(g,J)$ iff $Isubseteq J$ and $g|_I=f$.
Then there obviously exists some partial clusterpoint; consider $(f,emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $ain Asetminus I$. Then the $a$-coordinate of the subnet ${g_n|_I}$ of ${f_n|_I}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,Icup{a})$.
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
$endgroup$
add a comment |
$begingroup$
The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $omega_1$, the first uncountable ordinal, then there exists no final function $h:mathbb{N}toomega_1$, since $cup h(n)$ is an upper bound on $h(n)$.
To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence ${f_n}$ in $Pi X_{alphain A}$ as a tuple $(f,I)$ where $finPi X_{alphain A}$ and $Isubseteq A$ such that $f$ is a cluster point of ${f_n|_I}$. We can then define the partial ordening $(f,I)leq(g,J)$ iff $Isubseteq J$ and $g|_I=f$.
Then there obviously exists some partial clusterpoint; consider $(f,emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $ain Asetminus I$. Then the $a$-coordinate of the subnet ${g_n|_I}$ of ${f_n|_I}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,Icup{a})$.
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
$endgroup$
add a comment |
$begingroup$
The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $omega_1$, the first uncountable ordinal, then there exists no final function $h:mathbb{N}toomega_1$, since $cup h(n)$ is an upper bound on $h(n)$.
To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence ${f_n}$ in $Pi X_{alphain A}$ as a tuple $(f,I)$ where $finPi X_{alphain A}$ and $Isubseteq A$ such that $f$ is a cluster point of ${f_n|_I}$. We can then define the partial ordening $(f,I)leq(g,J)$ iff $Isubseteq J$ and $g|_I=f$.
Then there obviously exists some partial clusterpoint; consider $(f,emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $ain Asetminus I$. Then the $a$-coordinate of the subnet ${g_n|_I}$ of ${f_n|_I}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,Icup{a})$.
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
$endgroup$
The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $omega_1$, the first uncountable ordinal, then there exists no final function $h:mathbb{N}toomega_1$, since $cup h(n)$ is an upper bound on $h(n)$.
To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence ${f_n}$ in $Pi X_{alphain A}$ as a tuple $(f,I)$ where $finPi X_{alphain A}$ and $Isubseteq A$ such that $f$ is a cluster point of ${f_n|_I}$. We can then define the partial ordening $(f,I)leq(g,J)$ iff $Isubseteq J$ and $g|_I=f$.
Then there obviously exists some partial clusterpoint; consider $(f,emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $ain Asetminus I$. Then the $a$-coordinate of the subnet ${g_n|_I}$ of ${f_n|_I}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,Icup{a})$.
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
edited Jan 21 at 21:35
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
answered Jan 20 at 9:22
SmileyCraftSmileyCraft
3,476518
3,476518
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1
$begingroup$
Can you give an example of a subnet that is not a subsequence?
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– SmileyCraft
Jan 20 at 5:40
$begingroup$
But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets.
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– SmileyCraft
Jan 20 at 5:49
$begingroup$
That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in ${0,1}^{[0,1]}$ with no convergent subsequence is ${f_n}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet.
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– SmileyCraft
Jan 20 at 6:03
$begingroup$
No, a subnet need not have a subsubnet that is a subsequence of the original sequence.
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– David C. Ullrich
Jan 20 at 18:46