Mixing
Proving $(-1,1)^{mathbb{N}}$ is not open in the product topology of $mathbb{R}^{mathbb{N}}$
$begingroup$
Clarification: here $mathbb{R}^{mathbb{N}} = mathbb{R}times mathbb{R} times cdots$, i.e, countably many copies of $mathbb{R}$. $(-1,1)^{mathbb{N}}$ is completely analagous.
I don't want a proof using sequences, so here's what I've done:
Suppose $(-1,1)^{mathbb{N}}$ is open in the product topology. Then it contains a basis element of the product topology. Since any basis element of the product topology in this case is of the form $B = displaystyle{prod_{alpha in J}} U_alpha$, where $U_alpha = mathbb{R}$ for all but finitely many values of $alpha$, we have a contradiction, since obviously there are elements of $mathbb{R}$ that aren't coordinates of $(-1,1)^{mathbb{N}}$ (and, by definition, for a set to be open in the product topology, all of it's coordinates have to be in all of the $U_alpha$). Therefore $(-1,1)^{mathbb{N}}$ doesn't contain any element of the product topology and it follows that it's not open.
Now, have I done anything wrong here or made myself unclear? How could I improve what I wrote? Is it alright? I appreciate any help.
general-topology product-space box-topology
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
|
show 3 more comments
$begingroup$
Clarification: here $mathbb{R}^{mathbb{N}} = mathbb{R}times mathbb{R} times cdots$, i.e, countably many copies of $mathbb{R}$. $(-1,1)^{mathbb{N}}$ is completely analagous.
I don't want a proof using sequences, so here's what I've done:
Suppose $(-1,1)^{mathbb{N}}$ is open in the product topology. Then it contains a basis element of the product topology. Since any basis element of the product topology in this case is of the form $B = displaystyle{prod_{alpha in J}} U_alpha$, where $U_alpha = mathbb{R}$ for all but finitely many values of $alpha$, we have a contradiction, since obviously there are elements of $mathbb{R}$ that aren't coordinates of $(-1,1)^{mathbb{N}}$ (and, by definition, for a set to be open in the product topology, all of it's coordinates have to be in all of the $U_alpha$). Therefore $(-1,1)^{mathbb{N}}$ doesn't contain any element of the product topology and it follows that it's not open.
Now, have I done anything wrong here or made myself unclear? How could I improve what I wrote? Is it alright? I appreciate any help.
general-topology product-space box-topology
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
1
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
2
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
1
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
1
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50
|
show 3 more comments
$begingroup$
Clarification: here $mathbb{R}^{mathbb{N}} = mathbb{R}times mathbb{R} times cdots$, i.e, countably many copies of $mathbb{R}$. $(-1,1)^{mathbb{N}}$ is completely analagous.
I don't want a proof using sequences, so here's what I've done:
Suppose $(-1,1)^{mathbb{N}}$ is open in the product topology. Then it contains a basis element of the product topology. Since any basis element of the product topology in this case is of the form $B = displaystyle{prod_{alpha in J}} U_alpha$, where $U_alpha = mathbb{R}$ for all but finitely many values of $alpha$, we have a contradiction, since obviously there are elements of $mathbb{R}$ that aren't coordinates of $(-1,1)^{mathbb{N}}$ (and, by definition, for a set to be open in the product topology, all of it's coordinates have to be in all of the $U_alpha$). Therefore $(-1,1)^{mathbb{N}}$ doesn't contain any element of the product topology and it follows that it's not open.
Now, have I done anything wrong here or made myself unclear? How could I improve what I wrote? Is it alright? I appreciate any help.
general-topology product-space box-topology
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
Clarification: here $mathbb{R}^{mathbb{N}} = mathbb{R}times mathbb{R} times cdots$, i.e, countably many copies of $mathbb{R}$. $(-1,1)^{mathbb{N}}$ is completely analagous.
I don't want a proof using sequences, so here's what I've done:
Suppose $(-1,1)^{mathbb{N}}$ is open in the product topology. Then it contains a basis element of the product topology. Since any basis element of the product topology in this case is of the form $B = displaystyle{prod_{alpha in J}} U_alpha$, where $U_alpha = mathbb{R}$ for all but finitely many values of $alpha$, we have a contradiction, since obviously there are elements of $mathbb{R}$ that aren't coordinates of $(-1,1)^{mathbb{N}}$ (and, by definition, for a set to be open in the product topology, all of it's coordinates have to be in all of the $U_alpha$). Therefore $(-1,1)^{mathbb{N}}$ doesn't contain any element of the product topology and it follows that it's not open.
Now, have I done anything wrong here or made myself unclear? How could I improve what I wrote? Is it alright? I appreciate any help.
general-topology product-space box-topology
general-topology product-space box-topology
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
edited Jan 16 at 21:58
Matheus Andrade
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
asked Jan 16 at 20:40
Matheus AndradeMatheus Andrade
1,374418
1,374418
1
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
2
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
1
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
1
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50
|
show 3 more comments
1
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
2
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
1
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
1
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50
1
1
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
2
2
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
1
1
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
1
1
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50
|
show 3 more comments
1 Answer
1
active
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$begingroup$
The idea is correct as you wrote it: suppose that $$B=prod_{n in mathbb{N}} U_n subseteq (-1,1)^{mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $mathbb{N}$ such that $U_n = mathbb{R}$ for all $n notin F$, and all other $U_n$ non-empty open subsets of $mathbb{R}$.
Then the point $(p_n)$ defined by picking $p_nin U_n$ for $nin F$ and setting $p_n = 2$ for $n notin F$, obeys $p in B$ by construction, but $p notin (-1,1)^{mathbb{N}}$ as witnessed by any coordinate $n notin F$ $(p in (-1,1)^{mathbb{N}}$ iff $forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{mathbb{N}}$ is empty.
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
add a comment |
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1 Answer
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$begingroup$
The idea is correct as you wrote it: suppose that $$B=prod_{n in mathbb{N}} U_n subseteq (-1,1)^{mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $mathbb{N}$ such that $U_n = mathbb{R}$ for all $n notin F$, and all other $U_n$ non-empty open subsets of $mathbb{R}$.
Then the point $(p_n)$ defined by picking $p_nin U_n$ for $nin F$ and setting $p_n = 2$ for $n notin F$, obeys $p in B$ by construction, but $p notin (-1,1)^{mathbb{N}}$ as witnessed by any coordinate $n notin F$ $(p in (-1,1)^{mathbb{N}}$ iff $forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{mathbb{N}}$ is empty.
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
add a comment |
$begingroup$
The idea is correct as you wrote it: suppose that $$B=prod_{n in mathbb{N}} U_n subseteq (-1,1)^{mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $mathbb{N}$ such that $U_n = mathbb{R}$ for all $n notin F$, and all other $U_n$ non-empty open subsets of $mathbb{R}$.
Then the point $(p_n)$ defined by picking $p_nin U_n$ for $nin F$ and setting $p_n = 2$ for $n notin F$, obeys $p in B$ by construction, but $p notin (-1,1)^{mathbb{N}}$ as witnessed by any coordinate $n notin F$ $(p in (-1,1)^{mathbb{N}}$ iff $forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{mathbb{N}}$ is empty.
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
add a comment |
$begingroup$
The idea is correct as you wrote it: suppose that $$B=prod_{n in mathbb{N}} U_n subseteq (-1,1)^{mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $mathbb{N}$ such that $U_n = mathbb{R}$ for all $n notin F$, and all other $U_n$ non-empty open subsets of $mathbb{R}$.
Then the point $(p_n)$ defined by picking $p_nin U_n$ for $nin F$ and setting $p_n = 2$ for $n notin F$, obeys $p in B$ by construction, but $p notin (-1,1)^{mathbb{N}}$ as witnessed by any coordinate $n notin F$ $(p in (-1,1)^{mathbb{N}}$ iff $forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{mathbb{N}}$ is empty.
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
The idea is correct as you wrote it: suppose that $$B=prod_{n in mathbb{N}} U_n subseteq (-1,1)^{mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $mathbb{N}$ such that $U_n = mathbb{R}$ for all $n notin F$, and all other $U_n$ non-empty open subsets of $mathbb{R}$.
Then the point $(p_n)$ defined by picking $p_nin U_n$ for $nin F$ and setting $p_n = 2$ for $n notin F$, obeys $p in B$ by construction, but $p notin (-1,1)^{mathbb{N}}$ as witnessed by any coordinate $n notin F$ $(p in (-1,1)^{mathbb{N}}$ iff $forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{mathbb{N}}$ is empty.
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
answered Jan 16 at 21:57
Henno BrandsmaHenno Brandsma
110k348118
110k348118
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
add a comment |
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
$begingroup$
This is only my second week with General Topology so I took a while to understand your answer, but it absolutely could not be any clearer and I consider it a great improvement to my original proof since it actually builds on all the ideas I only mentioned. Thank you!
$endgroup$
– Matheus Andrade
Jan 16 at 22:17
add a comment |
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1
$begingroup$
An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element.
$endgroup$
– Robert Israel
Jan 16 at 20:43
$begingroup$
Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now?
$endgroup$
– Matheus Andrade
Jan 16 at 20:46
2
$begingroup$
@MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that.
$endgroup$
– Cameron Williams
Jan 16 at 20:54
1
$begingroup$
@MatheusAndrade I think about these typologies as follows: If you consider $mathbb{R}^mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $ninmathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $ninmathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better)
$endgroup$
– Yanko
Jan 16 at 21:03
1
$begingroup$
@Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent.
$endgroup$
– Henno Brandsma
Jan 16 at 21:50