Showing the closure of a compact subset need not be compactTopology: Example of a compact set but its closure...
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Showing the closure of a compact subset need not be compact
Topology: Example of a compact set but its closure not compactClosure of a compact space always 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
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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 yesterday
Austin Mohr
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
$endgroup$
|
show 3 more comments
$begingroup$
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 yesterday
Austin Mohr
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
$endgroup$
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
yesterday
|
show 3 more comments
$begingroup$
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 yesterday
Austin Mohr
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
$endgroup$
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 yesterday
Austin Mohr
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
edited yesterday
Austin Mohr
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
edited yesterday
Austin Mohr
20.8k35299
edited yesterday
Austin Mohr
20.8k35299
edited yesterday
Austin Mohr
20.8k35299
20.8k35299
asked yesterday
can'tcauchycan'tcauchy
1,022417
asked yesterday
can'tcauchycan'tcauchy
1,022417
asked yesterday
can'tcauchycan'tcauchy
1,022417
1,022417
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
yesterday
|
show 3 more comments
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
yesterday
|
show 3 more comments
2 Answers
2
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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 yesterday
parsiadparsiad
18.7k32453
$endgroup$
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
$begingroup$
@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
$endgroup$
– parsiad
13 hours ago
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 yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 yesterday
parsiadparsiad
18.7k32453
$endgroup$
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
$begingroup$
@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
$endgroup$
– parsiad
13 hours ago
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 yesterday
parsiadparsiad
18.7k32453
$endgroup$
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
$begingroup$
@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
$endgroup$
– parsiad
13 hours ago
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 yesterday
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 yesterday
parsiadparsiad
18.7k32453
answered yesterday
parsiadparsiad
18.7k32453
answered yesterday
parsiadparsiad
18.7k32453
answered yesterday
parsiadparsiad
18.7k32453
18.7k32453
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
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– Jeppe Stig Nielsen
20 hours ago
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@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
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– parsiad
13 hours ago
add a comment |
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
$begingroup$
@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
$endgroup$
– parsiad
13 hours ago
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
$begingroup$
Is there a topological space $Y$ that has a compact subset $Asubset Y$ with infinitely many points ($|A|=infty$), and still $overline{A} ne A$? Maybe even $overline{A}$ is noncompact like above.
$endgroup$
– Jeppe Stig Nielsen
20 hours ago
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@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
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– parsiad
13 hours ago
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@JeppeStigNielsen: You can modify the same idea in OP's example to get what you want :-)
$endgroup$
– parsiad
13 hours ago
add a comment |
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$A={1}$ 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 yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
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 yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
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 yesterday
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 yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
115k349125
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
add a comment |
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
$begingroup$
To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
$endgroup$
– DanielWainfleet
22 hours ago
add a comment |
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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
yesterday
$begingroup$
@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
yesterday
1
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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
yesterday
$begingroup$
@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 ?
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– can'tcauchy
yesterday
$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
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– parsiad
yesterday