Showing the closure of a compact subset need not be compactIsn't every subset of a compact space...
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Showing the closure of a compact subset need not be compact
Isn't every subset of a compact space compact?Example on closure of a subset of a subspace of a topological space in Munkres's TopologyCompact subset of a non compact topological spaceWhy subspace of a compact space not compactIs $mathbb{R}$ compact under the co-countable and co-finite topologies?Show that $mathbb{Q}$ is not locally compact with a characterization of local compactnessIs the closure of a compact set compact?A topological space is locally compact then here is an open base at each point has all of its set with compact closure$Bbb{R}^omega$ is not Locally CompactShowing a subset of $Bbb R^2$ is compact with the relative topology
$begingroup$
Could someone tell me if my line of reasoning is correct here:
Say we have the topological space $(mathbb{N}, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.
I want to show that the closure of a compact set in this topology need not be compact.
Let $A equiv {1, ...., n}$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overline{A} = Bbb N$
If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)
general-topology proof-writing compactness
edited 4 hours ago
Austin Mohr
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
$endgroup$
|
show 1 more comment
$begingroup$
Could someone tell me if my line of reasoning is correct here:
Say we have the topological space $(mathbb{N}, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.
I want to show that the closure of a compact set in this topology need not be compact.
Let $A equiv {1, ...., n}$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overline{A} = Bbb N$
If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)
general-topology proof-writing compactness
edited 4 hours ago
Austin Mohr
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
$endgroup$
3
$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago
1
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago
|
show 1 more comment
$begingroup$
Could someone tell me if my line of reasoning is correct here:
Say we have the topological space $(mathbb{N}, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.
I want to show that the closure of a compact set in this topology need not be compact.
Let $A equiv {1, ...., n}$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overline{A} = Bbb N$
If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)
general-topology proof-writing compactness
edited 4 hours ago
Austin Mohr
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
$endgroup$
Could someone tell me if my line of reasoning is correct here:
Say we have the topological space $(mathbb{N}, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.
I want to show that the closure of a compact set in this topology need not be compact.
Let $A equiv {1, ...., n}$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overline{A} = Bbb N$
If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)
general-topology proof-writing compactness
general-topology proof-writing compactness
edited 4 hours ago
Austin Mohr
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
edited 4 hours ago
Austin Mohr
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
edited 4 hours ago
Austin Mohr
20.8k35299
edited 4 hours ago
Austin Mohr
20.8k35299
edited 4 hours ago
Austin Mohr
20.8k35299
20.8k35299
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
asked 4 hours ago
can'tcauchycan'tcauchy
1,023417
1,023417
3
$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago
1
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago
|
show 1 more comment
3
$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago
1
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago
3
3
$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago
$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago
1
1
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection ${ B_alpha } subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection ${B_{alpha_1}, ldots, B_{alpha_n}}$ such that $B_{alpha_1 } cup cdots cup B_{alpha_n} supset A$.
Remark. We call the original collection an open cover and the subcollection a finite subcover.
Consider the topology in your original question.
Let $A equiv {1, ldots, n}$ and $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ is a cover of $A$, we can find $alpha_k$ such that $k in B_{alpha_k}$.
Therefore, ${B_{alpha_k}}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n equiv {1,ldots,n}$ and consider the cover $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overline{A} = mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
answered 4 hours ago
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
$begingroup$
$A={1}$ is compact as any cover of it has a one-element subcover.
$overline{A} = mathbb N$ which is not compact, as witnessed by the open cover $${{1,2},{1,3},{1,4},ldots, {1,n}, ldots}$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection ${ B_alpha } subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection ${B_{alpha_1}, ldots, B_{alpha_n}}$ such that $B_{alpha_1 } cup cdots cup B_{alpha_n} supset A$.
Remark. We call the original collection an open cover and the subcollection a finite subcover.
Consider the topology in your original question.
Let $A equiv {1, ldots, n}$ and $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ is a cover of $A$, we can find $alpha_k$ such that $k in B_{alpha_k}$.
Therefore, ${B_{alpha_k}}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n equiv {1,ldots,n}$ and consider the cover $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overline{A} = mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
answered 4 hours ago
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
$begingroup$
Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection ${ B_alpha } subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection ${B_{alpha_1}, ldots, B_{alpha_n}}$ such that $B_{alpha_1 } cup cdots cup B_{alpha_n} supset A$.
Remark. We call the original collection an open cover and the subcollection a finite subcover.
Consider the topology in your original question.
Let $A equiv {1, ldots, n}$ and $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ is a cover of $A$, we can find $alpha_k$ such that $k in B_{alpha_k}$.
Therefore, ${B_{alpha_k}}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n equiv {1,ldots,n}$ and consider the cover $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overline{A} = mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
answered 4 hours ago
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
$begingroup$
Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection ${ B_alpha } subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection ${B_{alpha_1}, ldots, B_{alpha_n}}$ such that $B_{alpha_1 } cup cdots cup B_{alpha_n} supset A$.
Remark. We call the original collection an open cover and the subcollection a finite subcover.
Consider the topology in your original question.
Let $A equiv {1, ldots, n}$ and $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ is a cover of $A$, we can find $alpha_k$ such that $k in B_{alpha_k}$.
Therefore, ${B_{alpha_k}}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n equiv {1,ldots,n}$ and consider the cover $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overline{A} = mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
answered 4 hours ago
parsiadparsiad
18.7k32453
$endgroup$
Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection ${ B_alpha } subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection ${B_{alpha_1}, ldots, B_{alpha_n}}$ such that $B_{alpha_1 } cup cdots cup B_{alpha_n} supset A$.
Remark. We call the original collection an open cover and the subcollection a finite subcover.
Consider the topology in your original question.
Let $A equiv {1, ldots, n}$ and $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ is a cover of $A$, we can find $alpha_k$ such that $k in B_{alpha_k}$.
Therefore, ${B_{alpha_k}}_{k=1}^n$ covers $A$, and hence $A$ is compact.
Next, let $C_n equiv {1,ldots,n}$ and consider the cover $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ is not compact.
Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overline{A} = mathbb{N}$.
In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.
answered 4 hours ago
parsiadparsiad
18.7k32453
answered 4 hours ago
parsiadparsiad
18.7k32453
answered 4 hours ago
parsiadparsiad
18.7k32453
answered 4 hours ago
parsiadparsiad
18.7k32453
18.7k32453
add a comment |
add a comment |
$begingroup$
$A={1}$ is compact as any cover of it has a one-element subcover.
$overline{A} = mathbb N$ which is not compact, as witnessed by the open cover $${{1,2},{1,3},{1,4},ldots, {1,n}, ldots}$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
$A={1}$ is compact as any cover of it has a one-element subcover.
$overline{A} = mathbb N$ which is not compact, as witnessed by the open cover $${{1,2},{1,3},{1,4},ldots, {1,n}, ldots}$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
$A={1}$ is compact as any cover of it has a one-element subcover.
$overline{A} = mathbb N$ which is not compact, as witnessed by the open cover $${{1,2},{1,3},{1,4},ldots, {1,n}, ldots}$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$A={1}$ is compact as any cover of it has a one-element subcover.
$overline{A} = mathbb N$ which is not compact, as witnessed by the open cover $${{1,2},{1,3},{1,4},ldots, {1,n}, ldots}$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
answered 1 hour ago
Henno BrandsmaHenno Brandsma
115k349125
115k349125
add a comment |
add a comment |
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$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = {1, ldots, n}$ is compact because given an open cover ${ B_alpha }$ of $A$, we can find $B_{alpha_k} ni k$ for each $k in A$ so that ${ B_{alpha_k} }_{k=1}^n$ covers $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
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– parsiad
4 hours ago
$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover ${B_alpha}$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
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– can'tcauchy
4 hours ago
1
$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
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– parsiad
4 hours ago
$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A={1,...n} as being able to be covered by the set {1,....n} in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago