Set of infinite subsets. Is it a topology?
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Here is the question:
Let $X = mathbb{R}$ and let $Omega$ consist of the empty set and all infinite subsets of $mathbb{R}$. Is $Omega$ a topological structure?
My attempt : I think the answer is No; it is not a topology.
Because we have $$ Omega = lbrace varnothing rbrace cup lbrace U subseteq mathbb{R} mid text{$U$ is infinite} rbrace $$
If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $Omega$ now $cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $Omega$. Is this explanation correct?
EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $Omega$ as they are infinite sets but $ U_1 cap U_2 = lbrace 2 rbrace $ which is finite.
general-topology
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
|
show 5 more comments
up vote
5
down vote
favorite
Here is the question:
Let $X = mathbb{R}$ and let $Omega$ consist of the empty set and all infinite subsets of $mathbb{R}$. Is $Omega$ a topological structure?
My attempt : I think the answer is No; it is not a topology.
Because we have $$ Omega = lbrace varnothing rbrace cup lbrace U subseteq mathbb{R} mid text{$U$ is infinite} rbrace $$
If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $Omega$ now $cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $Omega$. Is this explanation correct?
EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $Omega$ as they are infinite sets but $ U_1 cap U_2 = lbrace 2 rbrace $ which is finite.
general-topology
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
1
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
1
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00
|
show 5 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Here is the question:
Let $X = mathbb{R}$ and let $Omega$ consist of the empty set and all infinite subsets of $mathbb{R}$. Is $Omega$ a topological structure?
My attempt : I think the answer is No; it is not a topology.
Because we have $$ Omega = lbrace varnothing rbrace cup lbrace U subseteq mathbb{R} mid text{$U$ is infinite} rbrace $$
If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $Omega$ now $cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $Omega$. Is this explanation correct?
EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $Omega$ as they are infinite sets but $ U_1 cap U_2 = lbrace 2 rbrace $ which is finite.
general-topology
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
Here is the question:
Let $X = mathbb{R}$ and let $Omega$ consist of the empty set and all infinite subsets of $mathbb{R}$. Is $Omega$ a topological structure?
My attempt : I think the answer is No; it is not a topology.
Because we have $$ Omega = lbrace varnothing rbrace cup lbrace U subseteq mathbb{R} mid text{$U$ is infinite} rbrace $$
If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $Omega$ now $cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $Omega$. Is this explanation correct?
EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $Omega$ as they are infinite sets but $ U_1 cap U_2 = lbrace 2 rbrace $ which is finite.
general-topology
general-topology
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
edited Aug 12 '15 at 18:43
BunsOfWrath
630415
630415
asked Aug 12 '15 at 17:53
SMath
1,58811533
asked Aug 12 '15 at 17:53
SMath
1,58811533
asked Aug 12 '15 at 17:53
SMath
1,58811533
1,58811533
Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
1
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
1
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00
|
show 5 more comments
Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
1
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
1
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00
Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
1
1
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
1
1
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00
|
show 5 more comments
4 Answers
4
active
oldest
votes
up vote
2
down vote
You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just ${1}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
add a comment |
up vote
2
down vote
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is ${2}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$Omega={varnothing}cup{Usubseteq R,|,U^{mathsf c}text{ is finite}}.$$
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
add a comment |
up vote
-1
down vote
Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
answered Aug 12 '15 at 18:11
George
350212
add a comment |
up vote
-4
down vote
I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $mathbb{R}$ But it does not have $mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $mathbb{R}$ belong to taa but it have all infinite subsets of $mathbb{R}$, not $mathbb{R}$ itself. That's why in my view's taa is not a topology
edited Nov 21 at 15:56
MRobinson
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just ${1}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
add a comment |
up vote
2
down vote
You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just ${1}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
add a comment |
up vote
2
down vote
up vote
2
down vote
You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just ${1}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just ${1}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
edited Aug 12 '15 at 18:16
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
answered Aug 12 '15 at 18:11
user2566092
21.4k1946
21.4k1946
add a comment |
add a comment |
up vote
2
down vote
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is ${2}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$Omega={varnothing}cup{Usubseteq R,|,U^{mathsf c}text{ is finite}}.$$
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
add a comment |
up vote
2
down vote
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is ${2}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$Omega={varnothing}cup{Usubseteq R,|,U^{mathsf c}text{ is finite}}.$$
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
add a comment |
up vote
2
down vote
up vote
2
down vote
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is ${2}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$Omega={varnothing}cup{Usubseteq R,|,U^{mathsf c}text{ is finite}}.$$
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is ${2}$). This violates the condition that the intersection of two open sets be open in a topology.
However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$Omega={varnothing}cup{Usubseteq R,|,U^{mathsf c}text{ is finite}}.$$
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
edited Aug 12 '15 at 19:23
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
answered Aug 12 '15 at 18:50
triple_sec
15.7k21851
15.7k21851
add a comment |
add a comment |
up vote
-1
down vote
Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
answered Aug 12 '15 at 18:11
George
350212
add a comment |
up vote
-1
down vote
Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
answered Aug 12 '15 at 18:11
George
350212
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
answered Aug 12 '15 at 18:11
George
350212
Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..
answered Aug 12 '15 at 18:11
George
350212
edited Aug 12 '15 at 18:13
answered Aug 12 '15 at 18:11
George
350212
answered Aug 12 '15 at 18:11
George
350212
answered Aug 12 '15 at 18:11
George
350212
350212
add a comment |
add a comment |
up vote
-4
down vote
I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $mathbb{R}$ But it does not have $mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $mathbb{R}$ belong to taa but it have all infinite subsets of $mathbb{R}$, not $mathbb{R}$ itself. That's why in my view's taa is not a topology
edited Nov 21 at 15:56
MRobinson
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
add a comment |
up vote
-4
down vote
I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $mathbb{R}$ But it does not have $mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $mathbb{R}$ belong to taa but it have all infinite subsets of $mathbb{R}$, not $mathbb{R}$ itself. That's why in my view's taa is not a topology
edited Nov 21 at 15:56
MRobinson
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
add a comment |
up vote
-4
down vote
up vote
-4
down vote
I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $mathbb{R}$ But it does not have $mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $mathbb{R}$ belong to taa but it have all infinite subsets of $mathbb{R}$, not $mathbb{R}$ itself. That's why in my view's taa is not a topology
edited Nov 21 at 15:56
MRobinson
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $mathbb{R}$ But it does not have $mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $mathbb{R}$ belong to taa but it have all infinite subsets of $mathbb{R}$, not $mathbb{R}$ itself. That's why in my view's taa is not a topology
edited Nov 21 at 15:56
MRobinson
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
edited Nov 21 at 15:56
MRobinson
1,769319
edited Nov 21 at 15:56
MRobinson
1,769319
edited Nov 21 at 15:56
MRobinson
1,769319
1,769319
answered Nov 21 at 15:33
Arslan Sheikh
1
answered Nov 21 at 15:33
Arslan Sheikh
1
answered Nov 21 at 15:33
Arslan Sheikh
1
1
add a comment |
add a comment |
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Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$.
– mathcounterexamples.net
Aug 12 '15 at 17:56
1
“might be finite” needs evidence. Can you find two infinite subsets of $mathbb{R}$ with a finite intersection?
– Matthew Leingang
Aug 12 '15 at 17:57
You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets.
– Ian
Aug 12 '15 at 17:57
1
Does set of positive even numbers and primes work ?
– SMath
Aug 12 '15 at 17:59
Perhaps (it's your problem). But you don't need to use well-known sets.
– Matthew Leingang
Aug 12 '15 at 18:00