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Question on the Evenly Spaced Integer Topology
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Let $X$ be the set of integers with the topology generated by sets of the form $a + kmathbbZ = a+klambda, where $a,k in X$ and $k neq 0$. The basis sets are sipmly the cosets of nonzero subgoups of the integers.
This is from the book "Counterexamples in Topology" by Steen/Seebach. It goes on to say that points in $X$ are closed but not open, and i'm having trouble making sense of that.
I would like somebody to explain to me the two different reasons why each point is closed, i.e. I'm wondering:
How is the complement of a point a union of basis elements?
How do i know that a set containing just a single point contains all its limit points?
Thanks everyone, I have a feeling this is going to be one of those "oh duh" questions once somebody points out to me the reasoning.
general-topology
asked Jul 15 at 17:10
Michael Vaughan
29711
add a comment |Â
up vote
0
down vote
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Let $X$ be the set of integers with the topology generated by sets of the form $a + kmathbbZ = a+klambda, where $a,k in X$ and $k neq 0$. The basis sets are sipmly the cosets of nonzero subgoups of the integers.
This is from the book "Counterexamples in Topology" by Steen/Seebach. It goes on to say that points in $X$ are closed but not open, and i'm having trouble making sense of that.
I would like somebody to explain to me the two different reasons why each point is closed, i.e. I'm wondering:
How is the complement of a point a union of basis elements?
How do i know that a set containing just a single point contains all its limit points?
Thanks everyone, I have a feeling this is going to be one of those "oh duh" questions once somebody points out to me the reasoning.
general-topology
asked Jul 15 at 17:10
Michael Vaughan
29711
1
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
1
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be the set of integers with the topology generated by sets of the form $a + kmathbbZ = a+klambda, where $a,k in X$ and $k neq 0$. The basis sets are sipmly the cosets of nonzero subgoups of the integers.
This is from the book "Counterexamples in Topology" by Steen/Seebach. It goes on to say that points in $X$ are closed but not open, and i'm having trouble making sense of that.
I would like somebody to explain to me the two different reasons why each point is closed, i.e. I'm wondering:
How is the complement of a point a union of basis elements?
How do i know that a set containing just a single point contains all its limit points?
Thanks everyone, I have a feeling this is going to be one of those "oh duh" questions once somebody points out to me the reasoning.
general-topology
asked Jul 15 at 17:10
Michael Vaughan
29711
Let $X$ be the set of integers with the topology generated by sets of the form $a + kmathbbZ = a+klambda, where $a,k in X$ and $k neq 0$. The basis sets are sipmly the cosets of nonzero subgoups of the integers.
This is from the book "Counterexamples in Topology" by Steen/Seebach. It goes on to say that points in $X$ are closed but not open, and i'm having trouble making sense of that.
I would like somebody to explain to me the two different reasons why each point is closed, i.e. I'm wondering:
How is the complement of a point a union of basis elements?
How do i know that a set containing just a single point contains all its limit points?
Thanks everyone, I have a feeling this is going to be one of those "oh duh" questions once somebody points out to me the reasoning.
general-topology
asked Jul 15 at 17:10
Michael Vaughan
29711
asked Jul 15 at 17:10
Michael Vaughan
29711
asked Jul 15 at 17:10
Michael Vaughan
29711
asked Jul 15 at 17:10
Michael Vaughan
29711
29711
1
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
1
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30
add a comment |Â
1
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
1
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30
1
1
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
1
1
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The set $0$ is closed because $mathbbZsetminus0$ is the union of the following open sets:
- $1+2mathbb Z$;
- $2+4mathbb Z$;
- $4+8mathbb Z$;
- $cdots$
A similar argument applies to any other integer.
In order to see that if $nneq0$ then $n$ is not a limit point of $0$, just note that there are always integers $a$ and $k$, with $kneq0$, such that $0notin a+kmathbb Z$ and that $nin a+kmathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $nin a+kmathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $0$.
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
Given $x$, for every $yne x$ there exist $a$ and $kne 0$ such that $xnotin a+kBbb Z$ and $yin a+kBbb Z$. Just pick $k=|x-y|+1$ and $a=y$.
Hence the complement of $x$ is open.
Let $y$ be a limit point of a sequence living in $x$, i.e., of the constant sequence $x,x,ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)Bbb Z$, we find that $y=x$.
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff
(i). $cup B=X,$ and
(ii). if $b_1,b_2in B$ and $sin b_1cap b_2$ then there exists $b_3in B$ such that $sin b_3subset b_1cap b_2.$
If $B$ is a base for a topology on $X$ then any non-empty $tin T$ is the union of some members of $B$ so at least one $bin B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $tin T$ is infinite, and no $s$ can be open.
So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.
In your Q, if $a,a'in X$ and $aequiv a' mod k$ for every $kin Bbb Z$ $0$ then $a=a'.$ So, given $a in X,$ for each $a'in X$ $a$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $cup_ane a'in XB(a')$ is open , and its closed complement is $a.$
BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $Bbb Z$ is actually homeomorphic to the usual topology on $Bbb Q.$
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The set $0$ is closed because $mathbbZsetminus0$ is the union of the following open sets:
- $1+2mathbb Z$;
- $2+4mathbb Z$;
- $4+8mathbb Z$;
- $cdots$
A similar argument applies to any other integer.
In order to see that if $nneq0$ then $n$ is not a limit point of $0$, just note that there are always integers $a$ and $k$, with $kneq0$, such that $0notin a+kmathbb Z$ and that $nin a+kmathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $nin a+kmathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $0$.
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
add a comment |Â
up vote
2
down vote
accepted
The set $0$ is closed because $mathbbZsetminus0$ is the union of the following open sets:
- $1+2mathbb Z$;
- $2+4mathbb Z$;
- $4+8mathbb Z$;
- $cdots$
A similar argument applies to any other integer.
In order to see that if $nneq0$ then $n$ is not a limit point of $0$, just note that there are always integers $a$ and $k$, with $kneq0$, such that $0notin a+kmathbb Z$ and that $nin a+kmathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $nin a+kmathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $0$.
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The set $0$ is closed because $mathbbZsetminus0$ is the union of the following open sets:
- $1+2mathbb Z$;
- $2+4mathbb Z$;
- $4+8mathbb Z$;
- $cdots$
A similar argument applies to any other integer.
In order to see that if $nneq0$ then $n$ is not a limit point of $0$, just note that there are always integers $a$ and $k$, with $kneq0$, such that $0notin a+kmathbb Z$ and that $nin a+kmathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $nin a+kmathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $0$.
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
The set $0$ is closed because $mathbbZsetminus0$ is the union of the following open sets:
- $1+2mathbb Z$;
- $2+4mathbb Z$;
- $4+8mathbb Z$;
- $cdots$
A similar argument applies to any other integer.
In order to see that if $nneq0$ then $n$ is not a limit point of $0$, just note that there are always integers $a$ and $k$, with $kneq0$, such that $0notin a+kmathbb Z$ and that $nin a+kmathbb Z$; just take $a=n$ and $k=2n$, for instance. So, $nin a+kmathbb Z$ which is an open set to which $0$ does not bolong. Therefore, $n$ is not a limit point of $0$.
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
edited Jul 15 at 17:23
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
answered Jul 15 at 17:15
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
1
down vote
Given $x$, for every $yne x$ there exist $a$ and $kne 0$ such that $xnotin a+kBbb Z$ and $yin a+kBbb Z$. Just pick $k=|x-y|+1$ and $a=y$.
Hence the complement of $x$ is open.
Let $y$ be a limit point of a sequence living in $x$, i.e., of the constant sequence $x,x,ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)Bbb Z$, we find that $y=x$.
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
Given $x$, for every $yne x$ there exist $a$ and $kne 0$ such that $xnotin a+kBbb Z$ and $yin a+kBbb Z$. Just pick $k=|x-y|+1$ and $a=y$.
Hence the complement of $x$ is open.
Let $y$ be a limit point of a sequence living in $x$, i.e., of the constant sequence $x,x,ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)Bbb Z$, we find that $y=x$.
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given $x$, for every $yne x$ there exist $a$ and $kne 0$ such that $xnotin a+kBbb Z$ and $yin a+kBbb Z$. Just pick $k=|x-y|+1$ and $a=y$.
Hence the complement of $x$ is open.
Let $y$ be a limit point of a sequence living in $x$, i.e., of the constant sequence $x,x,ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)Bbb Z$, we find that $y=x$.
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
Given $x$, for every $yne x$ there exist $a$ and $kne 0$ such that $xnotin a+kBbb Z$ and $yin a+kBbb Z$. Just pick $k=|x-y|+1$ and $a=y$.
Hence the complement of $x$ is open.
Let $y$ be a limit point of a sequence living in $x$, i.e., of the constant sequence $x,x,ldots$. Then every neighbourhood of $y$ contains some term of that sequence, i.e., contains $x$. By considering the neighbourhood $y+(|x-y|+1)Bbb Z$, we find that $y=x$.
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
answered Jul 15 at 17:21
Hagen von Eitzen
265k20258477
265k20258477
add a comment |Â
add a comment |Â
up vote
1
down vote
A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff
(i). $cup B=X,$ and
(ii). if $b_1,b_2in B$ and $sin b_1cap b_2$ then there exists $b_3in B$ such that $sin b_3subset b_1cap b_2.$
If $B$ is a base for a topology on $X$ then any non-empty $tin T$ is the union of some members of $B$ so at least one $bin B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $tin T$ is infinite, and no $s$ can be open.
So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.
In your Q, if $a,a'in X$ and $aequiv a' mod k$ for every $kin Bbb Z$ $0$ then $a=a'.$ So, given $a in X,$ for each $a'in X$ $a$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $cup_ane a'in XB(a')$ is open , and its closed complement is $a.$
BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $Bbb Z$ is actually homeomorphic to the usual topology on $Bbb Q.$
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
add a comment |Â
up vote
1
down vote
A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff
(i). $cup B=X,$ and
(ii). if $b_1,b_2in B$ and $sin b_1cap b_2$ then there exists $b_3in B$ such that $sin b_3subset b_1cap b_2.$
If $B$ is a base for a topology on $X$ then any non-empty $tin T$ is the union of some members of $B$ so at least one $bin B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $tin T$ is infinite, and no $s$ can be open.
So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.
In your Q, if $a,a'in X$ and $aequiv a' mod k$ for every $kin Bbb Z$ $0$ then $a=a'.$ So, given $a in X,$ for each $a'in X$ $a$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $cup_ane a'in XB(a')$ is open , and its closed complement is $a.$
BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $Bbb Z$ is actually homeomorphic to the usual topology on $Bbb Q.$
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff
(i). $cup B=X,$ and
(ii). if $b_1,b_2in B$ and $sin b_1cap b_2$ then there exists $b_3in B$ such that $sin b_3subset b_1cap b_2.$
If $B$ is a base for a topology on $X$ then any non-empty $tin T$ is the union of some members of $B$ so at least one $bin B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $tin T$ is infinite, and no $s$ can be open.
So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.
In your Q, if $a,a'in X$ and $aequiv a' mod k$ for every $kin Bbb Z$ $0$ then $a=a'.$ So, given $a in X,$ for each $a'in X$ $a$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $cup_ane a'in XB(a')$ is open , and its closed complement is $a.$
BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $Bbb Z$ is actually homeomorphic to the usual topology on $Bbb Q.$
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
A family $B$ of subsets of a set $X$ is a base (basis) for a topology $T$ on $X$ iff
(i). $cup B=X,$ and
(ii). if $b_1,b_2in B$ and $sin b_1cap b_2$ then there exists $b_3in B$ such that $sin b_3subset b_1cap b_2.$
If $B$ is a base for a topology on $X$ then any non-empty $tin T$ is the union of some members of $B$ so at least one $bin B$ is a subset of $t.$ So if every member of $B$ is infinite then every non-empty $tin T$ is infinite, and no $s$ can be open.
So in your Q, once we have proved that what we have is indeed a base for a topology, it follows that all non-empty open sets are infinite.
In your Q, if $a,a'in X$ and $aequiv a' mod k$ for every $kin Bbb Z$ $0$ then $a=a'.$ So, given $a in X,$ for each $a'in X$ $a$ let $B(a')$ be a member of the base that contains $a'$ but not $a.$ Then $cup_ane a'in XB(a')$ is open , and its closed complement is $a.$
BTW. It is not hard to show that this topology is Hausdorff. Some deeper results imply that this topology on $Bbb Z$ is actually homeomorphic to the usual topology on $Bbb Q.$
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
answered Jul 15 at 18:09
DanielWainfleet
31.7k31644
31.7k31644
add a comment |Â
add a comment |Â
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1
I am curious as to what the book says this is a counter-example $to.$
– DanielWainfleet
Jul 15 at 18:10
Eh, it doesn't explicity say. It's a great book though, has 150 different interesting typologies and then does an analysis of each of them.
– Michael Vaughan
Jul 16 at 15:31
1
I'm a little familiar with this book. As Prof. Franklin Tall said, topology is known as much for its counter-example as its examples.
– DanielWainfleet
Jul 17 at 1:30