Mixing
Show that $Cl_betamathbbN0,2,4,…$ is open in $betamathbbN$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $betamathbbN$ denote the Stone-Cech compactification of the space $mathbbN$ of the natural numbers with the discrete topology and let $E$ denote the set of even natural numbers. Show that the closure $Cl_betamathbbN(E)$ is open in $betamathbbN$.
I understand the universal property of the Stone-Cech compactification. I understand that the Stone-Cech compactification is the maximal compactification and I understand the sense in which it is maximal. I also understand the standard construction of the Stone-Cech compactification as the closure of the image of an imbedding of your topological space into $[0,1]^C$ where $C$ is the set of continuous function from your space to $[0,1]$. However, I don't know how to determine when a subset of the Stone-Cech compactification is open nor do I know what any of the elements of $beta Xsetminus X$ look like for a space $X$.
general-topology compactification
asked Aug 6 at 16:25
Julian Benali
27212
add a comment |Â
up vote
1
down vote
favorite
Let $betamathbbN$ denote the Stone-Cech compactification of the space $mathbbN$ of the natural numbers with the discrete topology and let $E$ denote the set of even natural numbers. Show that the closure $Cl_betamathbbN(E)$ is open in $betamathbbN$.
I understand the universal property of the Stone-Cech compactification. I understand that the Stone-Cech compactification is the maximal compactification and I understand the sense in which it is maximal. I also understand the standard construction of the Stone-Cech compactification as the closure of the image of an imbedding of your topological space into $[0,1]^C$ where $C$ is the set of continuous function from your space to $[0,1]$. However, I don't know how to determine when a subset of the Stone-Cech compactification is open nor do I know what any of the elements of $beta Xsetminus X$ look like for a space $X$.
general-topology compactification
asked Aug 6 at 16:25
Julian Benali
27212
3
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
1
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $betamathbbN$ denote the Stone-Cech compactification of the space $mathbbN$ of the natural numbers with the discrete topology and let $E$ denote the set of even natural numbers. Show that the closure $Cl_betamathbbN(E)$ is open in $betamathbbN$.
I understand the universal property of the Stone-Cech compactification. I understand that the Stone-Cech compactification is the maximal compactification and I understand the sense in which it is maximal. I also understand the standard construction of the Stone-Cech compactification as the closure of the image of an imbedding of your topological space into $[0,1]^C$ where $C$ is the set of continuous function from your space to $[0,1]$. However, I don't know how to determine when a subset of the Stone-Cech compactification is open nor do I know what any of the elements of $beta Xsetminus X$ look like for a space $X$.
general-topology compactification
asked Aug 6 at 16:25
Julian Benali
27212
Let $betamathbbN$ denote the Stone-Cech compactification of the space $mathbbN$ of the natural numbers with the discrete topology and let $E$ denote the set of even natural numbers. Show that the closure $Cl_betamathbbN(E)$ is open in $betamathbbN$.
I understand the universal property of the Stone-Cech compactification. I understand that the Stone-Cech compactification is the maximal compactification and I understand the sense in which it is maximal. I also understand the standard construction of the Stone-Cech compactification as the closure of the image of an imbedding of your topological space into $[0,1]^C$ where $C$ is the set of continuous function from your space to $[0,1]$. However, I don't know how to determine when a subset of the Stone-Cech compactification is open nor do I know what any of the elements of $beta Xsetminus X$ look like for a space $X$.
general-topology compactification
asked Aug 6 at 16:25
Julian Benali
27212
edited Aug 8 at 16:25
asked Aug 6 at 16:25
Julian Benali
27212
asked Aug 6 at 16:25
Julian Benali
27212
asked Aug 6 at 16:25
Julian Benali
27212
27212
3
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
1
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38
add a comment |Â
3
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
1
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38
3
3
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
1
1
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Hint: Consider the "parity" function $f:mathbbNtoleft0,1right$,
$$f(x)=begincases0&text if xtext is even\
1&text if xtext is oddendcases$$
and remember that $left0,1right$ is compact.
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
add a comment |Â
up vote
0
down vote
accepted
Attempted solution to my problem:
I claim that $Cl_betamathbbN(E)$ and $Cl_betamathbbN(mathbbNsetminus E)$ are compliments of one another in $betamathbbN$. First, note that because $EsubsetmathbbN$, $Cl_betamathbbNE=Cl_[0,1]^C(E)capbetamathbbN=Cl_[0,1]^C(E)cap Cl_[0,1]^C(mathbbN)=Cl_[0,1]^C(E)$ which I will simply denote $overlineE$, and similarly $Cl_betamathbbN(mathbbNsetminus E)=Cl_[0,1]^C(mathbbNsetminus E)=overlinemathbbNsetminus E$.
Assume that $(x_f)_fin CinoverlineEcapoverlinemathbbNsetminus E$ and consider the parity function $gin C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_gneq 0$, then the open set $prod_fin CU_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $fneq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $E$ contradicting the assumption that $(x_f)_fin CinoverlineE$, so we must have $x_g=0$. But now, the open set $prod_fin CV_f$ defined by $V_f=[0,1]$ if $fneq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $mathbbNsetminus E$ contradicting the assumption that $(x_f)_fin CinoverlinemathbbNsetminus E$. Thus, by way of contradiction, it must be the case that $overlineEcapoverlinemathbbNsetminus E=emptyset$.
Finally, we see that $overlineEcupoverlinemathbbNsetminus E=overlineEcup(mathbbNsetminus E)=overlinemathbbN=betamathbbN$. Thus, we conclude that $overlineE$ and $overlinemathbbNsetminus E$ are compliments of one another. Since they are both closed, they are also both open.
answered Aug 11 at 20:15
Julian Benali
27212
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: Consider the "parity" function $f:mathbbNtoleft0,1right$,
$$f(x)=begincases0&text if xtext is even\
1&text if xtext is oddendcases$$
and remember that $left0,1right$ is compact.
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
add a comment |Â
up vote
2
down vote
Hint: Consider the "parity" function $f:mathbbNtoleft0,1right$,
$$f(x)=begincases0&text if xtext is even\
1&text if xtext is oddendcases$$
and remember that $left0,1right$ is compact.
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Consider the "parity" function $f:mathbbNtoleft0,1right$,
$$f(x)=begincases0&text if xtext is even\
1&text if xtext is oddendcases$$
and remember that $left0,1right$ is compact.
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
Hint: Consider the "parity" function $f:mathbbNtoleft0,1right$,
$$f(x)=begincases0&text if xtext is even\
1&text if xtext is oddendcases$$
and remember that $left0,1right$ is compact.
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
answered Aug 6 at 21:09
Luiz Cordeiro
12.2k1041
12.2k1041
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
add a comment |Â
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
Thanks. I don't see where the compactness of $0,1$ comes into play.
– Julian Benali
Aug 11 at 20:16
1
1
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
@JulianBenali because you can extend $f$ because of that compactness of the codomain.
– Henno Brandsma
Aug 11 at 20:45
1
1
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
Oh I see. Using the universal property of $betamathbbN$, $f$ extends to the function which sends $Cl_betamathbbNE$ to $0$ and its compliment to $1$. Thanks!
– Julian Benali
Aug 12 at 0:19
1
1
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
@JulianBenali Precisely. Note that this argument doesn't use any properties of $E$, and generalizes as follows: if $E$ is any clopen set of a locally compact Hausdorff $X$, then the closure of $E$ in $beta X$ is clopen.
– Luiz Cordeiro
Aug 13 at 0:18
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
I noticed that! Thank you!
– Julian Benali
Aug 14 at 0:17
add a comment |Â
up vote
0
down vote
accepted
Attempted solution to my problem:
I claim that $Cl_betamathbbN(E)$ and $Cl_betamathbbN(mathbbNsetminus E)$ are compliments of one another in $betamathbbN$. First, note that because $EsubsetmathbbN$, $Cl_betamathbbNE=Cl_[0,1]^C(E)capbetamathbbN=Cl_[0,1]^C(E)cap Cl_[0,1]^C(mathbbN)=Cl_[0,1]^C(E)$ which I will simply denote $overlineE$, and similarly $Cl_betamathbbN(mathbbNsetminus E)=Cl_[0,1]^C(mathbbNsetminus E)=overlinemathbbNsetminus E$.
Assume that $(x_f)_fin CinoverlineEcapoverlinemathbbNsetminus E$ and consider the parity function $gin C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_gneq 0$, then the open set $prod_fin CU_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $fneq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $E$ contradicting the assumption that $(x_f)_fin CinoverlineE$, so we must have $x_g=0$. But now, the open set $prod_fin CV_f$ defined by $V_f=[0,1]$ if $fneq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $mathbbNsetminus E$ contradicting the assumption that $(x_f)_fin CinoverlinemathbbNsetminus E$. Thus, by way of contradiction, it must be the case that $overlineEcapoverlinemathbbNsetminus E=emptyset$.
Finally, we see that $overlineEcupoverlinemathbbNsetminus E=overlineEcup(mathbbNsetminus E)=overlinemathbbN=betamathbbN$. Thus, we conclude that $overlineE$ and $overlinemathbbNsetminus E$ are compliments of one another. Since they are both closed, they are also both open.
answered Aug 11 at 20:15
Julian Benali
27212
add a comment |Â
up vote
0
down vote
accepted
Attempted solution to my problem:
I claim that $Cl_betamathbbN(E)$ and $Cl_betamathbbN(mathbbNsetminus E)$ are compliments of one another in $betamathbbN$. First, note that because $EsubsetmathbbN$, $Cl_betamathbbNE=Cl_[0,1]^C(E)capbetamathbbN=Cl_[0,1]^C(E)cap Cl_[0,1]^C(mathbbN)=Cl_[0,1]^C(E)$ which I will simply denote $overlineE$, and similarly $Cl_betamathbbN(mathbbNsetminus E)=Cl_[0,1]^C(mathbbNsetminus E)=overlinemathbbNsetminus E$.
Assume that $(x_f)_fin CinoverlineEcapoverlinemathbbNsetminus E$ and consider the parity function $gin C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_gneq 0$, then the open set $prod_fin CU_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $fneq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $E$ contradicting the assumption that $(x_f)_fin CinoverlineE$, so we must have $x_g=0$. But now, the open set $prod_fin CV_f$ defined by $V_f=[0,1]$ if $fneq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $mathbbNsetminus E$ contradicting the assumption that $(x_f)_fin CinoverlinemathbbNsetminus E$. Thus, by way of contradiction, it must be the case that $overlineEcapoverlinemathbbNsetminus E=emptyset$.
Finally, we see that $overlineEcupoverlinemathbbNsetminus E=overlineEcup(mathbbNsetminus E)=overlinemathbbN=betamathbbN$. Thus, we conclude that $overlineE$ and $overlinemathbbNsetminus E$ are compliments of one another. Since they are both closed, they are also both open.
answered Aug 11 at 20:15
Julian Benali
27212
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Attempted solution to my problem:
I claim that $Cl_betamathbbN(E)$ and $Cl_betamathbbN(mathbbNsetminus E)$ are compliments of one another in $betamathbbN$. First, note that because $EsubsetmathbbN$, $Cl_betamathbbNE=Cl_[0,1]^C(E)capbetamathbbN=Cl_[0,1]^C(E)cap Cl_[0,1]^C(mathbbN)=Cl_[0,1]^C(E)$ which I will simply denote $overlineE$, and similarly $Cl_betamathbbN(mathbbNsetminus E)=Cl_[0,1]^C(mathbbNsetminus E)=overlinemathbbNsetminus E$.
Assume that $(x_f)_fin CinoverlineEcapoverlinemathbbNsetminus E$ and consider the parity function $gin C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_gneq 0$, then the open set $prod_fin CU_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $fneq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $E$ contradicting the assumption that $(x_f)_fin CinoverlineE$, so we must have $x_g=0$. But now, the open set $prod_fin CV_f$ defined by $V_f=[0,1]$ if $fneq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $mathbbNsetminus E$ contradicting the assumption that $(x_f)_fin CinoverlinemathbbNsetminus E$. Thus, by way of contradiction, it must be the case that $overlineEcapoverlinemathbbNsetminus E=emptyset$.
Finally, we see that $overlineEcupoverlinemathbbNsetminus E=overlineEcup(mathbbNsetminus E)=overlinemathbbN=betamathbbN$. Thus, we conclude that $overlineE$ and $overlinemathbbNsetminus E$ are compliments of one another. Since they are both closed, they are also both open.
answered Aug 11 at 20:15
Julian Benali
27212
Attempted solution to my problem:
I claim that $Cl_betamathbbN(E)$ and $Cl_betamathbbN(mathbbNsetminus E)$ are compliments of one another in $betamathbbN$. First, note that because $EsubsetmathbbN$, $Cl_betamathbbNE=Cl_[0,1]^C(E)capbetamathbbN=Cl_[0,1]^C(E)cap Cl_[0,1]^C(mathbbN)=Cl_[0,1]^C(E)$ which I will simply denote $overlineE$, and similarly $Cl_betamathbbN(mathbbNsetminus E)=Cl_[0,1]^C(mathbbNsetminus E)=overlinemathbbNsetminus E$.
Assume that $(x_f)_fin CinoverlineEcapoverlinemathbbNsetminus E$ and consider the parity function $gin C$ which sends even natural numbers to $0$ and odd natural numbers to $1$. If $x_gneq 0$, then the open set $prod_fin CU_f$ of $[0,1]^C$ defined by $U_f=[0,1]$ if $fneq g$ and $U_g=(0,1]$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $E$ contradicting the assumption that $(x_f)_fin CinoverlineE$, so we must have $x_g=0$. But now, the open set $prod_fin CV_f$ defined by $V_f=[0,1]$ if $fneq g$ and $V_g=[0,1)$ is a neighborhood of $(x_f)_fin C$ which is disjoint from $mathbbNsetminus E$ contradicting the assumption that $(x_f)_fin CinoverlinemathbbNsetminus E$. Thus, by way of contradiction, it must be the case that $overlineEcapoverlinemathbbNsetminus E=emptyset$.
Finally, we see that $overlineEcupoverlinemathbbNsetminus E=overlineEcup(mathbbNsetminus E)=overlinemathbbN=betamathbbN$. Thus, we conclude that $overlineE$ and $overlinemathbbNsetminus E$ are compliments of one another. Since they are both closed, they are also both open.
answered Aug 11 at 20:15
Julian Benali
27212
answered Aug 11 at 20:15
Julian Benali
27212
answered Aug 11 at 20:15
Julian Benali
27212
answered Aug 11 at 20:15
Julian Benali
27212
27212
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874059%2fshow-that-cl-beta-mathbbn-0-2-4-is-open-in-beta-mathbbn%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Maybe show that its complement is $Cl_betamathbbN1,3,5,…$, obviously closed.
– GEdgar
Aug 6 at 16:54
1
For the description of $beta Xsetminus X$ (or $beta X$ more generally) when $X$ is discrete (e.g. $mathbbN$), I like the construction as the set of ultrafilters better
– Max
Aug 6 at 18:35
@GEdgar I honestly would have never guessed that those two sets were complements of one another.
– Julian Benali
Aug 11 at 19:38