Mixing
Proof verification that $overline{mathbb{R}^{infty}} = mathbb{R^omega}$ and $overline{mathbb{R}^{infty}} =...
$begingroup$
I want to rid myself of any misunderstandings I have, so feel free to nitpick my attempt all you want so that in the end it's as clear as possible!
Now, this is the exercise:
begin{array} { l } { text { Let } mathbb { R } ^ { infty } text { be the subset of } mathbb { R } ^ { omega } text { consisting of all sequences that are "eventually zero," } } \ { text { that is, all sequences } left( x _ { 1 } , x _ { 2 } , ldots right) text { such that } x _ { i } neq 0 text { for only finitely many values } } \ { text { of } i . text { What is the closure of } mathbb { R } ^ { omega } text { in } mathbb { R } ^ { omega } text { in the box and product topologies? Justify } } \ { text { your answer. } } end{array}
I will first show that $overline{mathbb{R}^{infty}} = mathbb{R^omega}$ in the product topology. Indeed, take any $x in mathbb{R^omega}$ and let $U$ be an open set of $mathbb{R^omega}$ such that $x in U$. Then $U$ contains a basis element $B$ of the product topology such that: $$x in B = displaystyle{prod_{n in mathbb{N}} U_{n}} subset U $$
and $U_n = mathbb{R} forall n in mathbb{N}setminus F$ where $F$ is a finite subset of $mathbb{N}$. Now, the sequence $(y_n)_{n in mathbb{N}}$ defined by:
$$begin{align*} &y_n = pi_n(x) forall n in F \ &y_n = 0 forall n notin Fend{align*}$$
is an element of $B$, by construction. It follows that $(y_n)_{n in mathbb{N}} in B subset U$, and also by construction we have that $(y_n)_{n in mathbb{N}} in mathbb{R}^{infty}$. Then: $$U bigcap mathbb{R}^infty neq emptyset$$
and therefore $overline{mathbb{R}^{infty}} = mathbb{R^omega}$.
$\$
Now I will show that $overline{mathbb{R}^{infty}} = mathbb{R^infty}$ in the box topology. For this, it suffices to show that any $x notin mathbb{R}^{infty}$ is not in $overline{mathbb{R}^{infty}}$, and this can be done by showing that there exists an open set $U ni x$ in the product topology such that $U bigcap mathbb{R}^{infty} = emptyset$. Indeed, by definition if $x = (x_n)_{n in mathbb{N}} notin mathbb{R}^{infty}$, then there exists an infinite set $I subset mathbb{N}$ such that $x_i neq 0 forall i in I$. Now, for: $$U = displaystyle{prod_{i in mathbb{N}} U_{i} }$$
where $U_i = mathbb{R} setminus {0} forall i in I$ and $U_i = mathbb{R}$ otherwise. As desired, it's clear that $x in U$ and $U bigcap mathbb{R}^{infty} = emptyset$ and we're done.
Have I made any unnecessary or not entirely correct steps here? Is there anything I should make clearer?
EDIT: As I thought, there were some things that could be (and indeed were) improved. Thanks a lot, Brevan and Henno! This kind of thinking is very important and I will always try to bear it in mind so that my proofs are always as clean and clear as possible.
general-topology proof-verification
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
add a comment |
$begingroup$
I want to rid myself of any misunderstandings I have, so feel free to nitpick my attempt all you want so that in the end it's as clear as possible!
Now, this is the exercise:
begin{array} { l } { text { Let } mathbb { R } ^ { infty } text { be the subset of } mathbb { R } ^ { omega } text { consisting of all sequences that are "eventually zero," } } \ { text { that is, all sequences } left( x _ { 1 } , x _ { 2 } , ldots right) text { such that } x _ { i } neq 0 text { for only finitely many values } } \ { text { of } i . text { What is the closure of } mathbb { R } ^ { omega } text { in } mathbb { R } ^ { omega } text { in the box and product topologies? Justify } } \ { text { your answer. } } end{array}
I will first show that $overline{mathbb{R}^{infty}} = mathbb{R^omega}$ in the product topology. Indeed, take any $x in mathbb{R^omega}$ and let $U$ be an open set of $mathbb{R^omega}$ such that $x in U$. Then $U$ contains a basis element $B$ of the product topology such that: $$x in B = displaystyle{prod_{n in mathbb{N}} U_{n}} subset U $$
and $U_n = mathbb{R} forall n in mathbb{N}setminus F$ where $F$ is a finite subset of $mathbb{N}$. Now, the sequence $(y_n)_{n in mathbb{N}}$ defined by:
$$begin{align*} &y_n = pi_n(x) forall n in F \ &y_n = 0 forall n notin Fend{align*}$$
is an element of $B$, by construction. It follows that $(y_n)_{n in mathbb{N}} in B subset U$, and also by construction we have that $(y_n)_{n in mathbb{N}} in mathbb{R}^{infty}$. Then: $$U bigcap mathbb{R}^infty neq emptyset$$
and therefore $overline{mathbb{R}^{infty}} = mathbb{R^omega}$.
$\$
Now I will show that $overline{mathbb{R}^{infty}} = mathbb{R^infty}$ in the box topology. For this, it suffices to show that any $x notin mathbb{R}^{infty}$ is not in $overline{mathbb{R}^{infty}}$, and this can be done by showing that there exists an open set $U ni x$ in the product topology such that $U bigcap mathbb{R}^{infty} = emptyset$. Indeed, by definition if $x = (x_n)_{n in mathbb{N}} notin mathbb{R}^{infty}$, then there exists an infinite set $I subset mathbb{N}$ such that $x_i neq 0 forall i in I$. Now, for: $$U = displaystyle{prod_{i in mathbb{N}} U_{i} }$$
where $U_i = mathbb{R} setminus {0} forall i in I$ and $U_i = mathbb{R}$ otherwise. As desired, it's clear that $x in U$ and $U bigcap mathbb{R}^{infty} = emptyset$ and we're done.
Have I made any unnecessary or not entirely correct steps here? Is there anything I should make clearer?
EDIT: As I thought, there were some things that could be (and indeed were) improved. Thanks a lot, Brevan and Henno! This kind of thinking is very important and I will always try to bear it in mind so that my proofs are always as clean and clear as possible.
general-topology proof-verification
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
1
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
2
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37
add a comment |
$begingroup$
I want to rid myself of any misunderstandings I have, so feel free to nitpick my attempt all you want so that in the end it's as clear as possible!
Now, this is the exercise:
begin{array} { l } { text { Let } mathbb { R } ^ { infty } text { be the subset of } mathbb { R } ^ { omega } text { consisting of all sequences that are "eventually zero," } } \ { text { that is, all sequences } left( x _ { 1 } , x _ { 2 } , ldots right) text { such that } x _ { i } neq 0 text { for only finitely many values } } \ { text { of } i . text { What is the closure of } mathbb { R } ^ { omega } text { in } mathbb { R } ^ { omega } text { in the box and product topologies? Justify } } \ { text { your answer. } } end{array}
I will first show that $overline{mathbb{R}^{infty}} = mathbb{R^omega}$ in the product topology. Indeed, take any $x in mathbb{R^omega}$ and let $U$ be an open set of $mathbb{R^omega}$ such that $x in U$. Then $U$ contains a basis element $B$ of the product topology such that: $$x in B = displaystyle{prod_{n in mathbb{N}} U_{n}} subset U $$
and $U_n = mathbb{R} forall n in mathbb{N}setminus F$ where $F$ is a finite subset of $mathbb{N}$. Now, the sequence $(y_n)_{n in mathbb{N}}$ defined by:
$$begin{align*} &y_n = pi_n(x) forall n in F \ &y_n = 0 forall n notin Fend{align*}$$
is an element of $B$, by construction. It follows that $(y_n)_{n in mathbb{N}} in B subset U$, and also by construction we have that $(y_n)_{n in mathbb{N}} in mathbb{R}^{infty}$. Then: $$U bigcap mathbb{R}^infty neq emptyset$$
and therefore $overline{mathbb{R}^{infty}} = mathbb{R^omega}$.
$\$
Now I will show that $overline{mathbb{R}^{infty}} = mathbb{R^infty}$ in the box topology. For this, it suffices to show that any $x notin mathbb{R}^{infty}$ is not in $overline{mathbb{R}^{infty}}$, and this can be done by showing that there exists an open set $U ni x$ in the product topology such that $U bigcap mathbb{R}^{infty} = emptyset$. Indeed, by definition if $x = (x_n)_{n in mathbb{N}} notin mathbb{R}^{infty}$, then there exists an infinite set $I subset mathbb{N}$ such that $x_i neq 0 forall i in I$. Now, for: $$U = displaystyle{prod_{i in mathbb{N}} U_{i} }$$
where $U_i = mathbb{R} setminus {0} forall i in I$ and $U_i = mathbb{R}$ otherwise. As desired, it's clear that $x in U$ and $U bigcap mathbb{R}^{infty} = emptyset$ and we're done.
Have I made any unnecessary or not entirely correct steps here? Is there anything I should make clearer?
EDIT: As I thought, there were some things that could be (and indeed were) improved. Thanks a lot, Brevan and Henno! This kind of thinking is very important and I will always try to bear it in mind so that my proofs are always as clean and clear as possible.
general-topology proof-verification
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
$endgroup$
I want to rid myself of any misunderstandings I have, so feel free to nitpick my attempt all you want so that in the end it's as clear as possible!
Now, this is the exercise:
begin{array} { l } { text { Let } mathbb { R } ^ { infty } text { be the subset of } mathbb { R } ^ { omega } text { consisting of all sequences that are "eventually zero," } } \ { text { that is, all sequences } left( x _ { 1 } , x _ { 2 } , ldots right) text { such that } x _ { i } neq 0 text { for only finitely many values } } \ { text { of } i . text { What is the closure of } mathbb { R } ^ { omega } text { in } mathbb { R } ^ { omega } text { in the box and product topologies? Justify } } \ { text { your answer. } } end{array}
I will first show that $overline{mathbb{R}^{infty}} = mathbb{R^omega}$ in the product topology. Indeed, take any $x in mathbb{R^omega}$ and let $U$ be an open set of $mathbb{R^omega}$ such that $x in U$. Then $U$ contains a basis element $B$ of the product topology such that: $$x in B = displaystyle{prod_{n in mathbb{N}} U_{n}} subset U $$
and $U_n = mathbb{R} forall n in mathbb{N}setminus F$ where $F$ is a finite subset of $mathbb{N}$. Now, the sequence $(y_n)_{n in mathbb{N}}$ defined by:
$$begin{align*} &y_n = pi_n(x) forall n in F \ &y_n = 0 forall n notin Fend{align*}$$
is an element of $B$, by construction. It follows that $(y_n)_{n in mathbb{N}} in B subset U$, and also by construction we have that $(y_n)_{n in mathbb{N}} in mathbb{R}^{infty}$. Then: $$U bigcap mathbb{R}^infty neq emptyset$$
and therefore $overline{mathbb{R}^{infty}} = mathbb{R^omega}$.
$\$
Now I will show that $overline{mathbb{R}^{infty}} = mathbb{R^infty}$ in the box topology. For this, it suffices to show that any $x notin mathbb{R}^{infty}$ is not in $overline{mathbb{R}^{infty}}$, and this can be done by showing that there exists an open set $U ni x$ in the product topology such that $U bigcap mathbb{R}^{infty} = emptyset$. Indeed, by definition if $x = (x_n)_{n in mathbb{N}} notin mathbb{R}^{infty}$, then there exists an infinite set $I subset mathbb{N}$ such that $x_i neq 0 forall i in I$. Now, for: $$U = displaystyle{prod_{i in mathbb{N}} U_{i} }$$
where $U_i = mathbb{R} setminus {0} forall i in I$ and $U_i = mathbb{R}$ otherwise. As desired, it's clear that $x in U$ and $U bigcap mathbb{R}^{infty} = emptyset$ and we're done.
Have I made any unnecessary or not entirely correct steps here? Is there anything I should make clearer?
EDIT: As I thought, there were some things that could be (and indeed were) improved. Thanks a lot, Brevan and Henno! This kind of thinking is very important and I will always try to bear it in mind so that my proofs are always as clean and clear as possible.
general-topology proof-verification
general-topology proof-verification
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
edited Jan 21 at 14:19
Matheus Andrade
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
asked Jan 17 at 22:30
Matheus AndradeMatheus Andrade
1,374418
1,374418
1
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
2
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37
add a comment |
1
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
2
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37
1
1
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
2
2
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37
add a comment |
1 Answer
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$begingroup$
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B cap mathbb{R}^inftyneq emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $mathbb{R}setminus{0}$ explicitly for those non-zero coordinates, and (like you did) $mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
add a comment |
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$begingroup$
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B cap mathbb{R}^inftyneq emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $mathbb{R}setminus{0}$ explicitly for those non-zero coordinates, and (like you did) $mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
add a comment |
$begingroup$
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B cap mathbb{R}^inftyneq emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $mathbb{R}setminus{0}$ explicitly for those non-zero coordinates, and (like you did) $mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
add a comment |
$begingroup$
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B cap mathbb{R}^inftyneq emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $mathbb{R}setminus{0}$ explicitly for those non-zero coordinates, and (like you did) $mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B cap mathbb{R}^inftyneq emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $mathbb{R}setminus{0}$ explicitly for those non-zero coordinates, and (like you did) $mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
edited Jan 17 at 22:51
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 17 at 22:46
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
add a comment |
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
$begingroup$
Thank you! This has helped a lot and made me aware of some things I definitely should always bear in mind.
$endgroup$
– Matheus Andrade
Jan 17 at 23:38
add a comment |
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1
$begingroup$
First thing that sticks out to me is that you could be a bit more clear on how you get the sequence $x_n.$ It of course suffices to simply let $x_n$ the $n$th coordinate of $x$ for all $n in F,$ and to let $x_n = 0$ otherwise.
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:42
2
$begingroup$
Everything is definitely correct. The subsequence notation at the end gets a bit harder to read, and you might just consider saying that $x_i neq 0$ for all $i$ in some index set $I,$ so that you can simply consider $(mathbb{N}setminus I) cup I$
$endgroup$
– Brevan Ellefsen
Jan 17 at 22:49
$begingroup$
@BrevanEllefsen Thanks! That does indeed make things better.
$endgroup$
– Matheus Andrade
Jan 17 at 23:37