Mixing
Properties of this topology on $Bbb X$.
$begingroup$
For given the usual topology
$tau$ on $Bbb{R}$, define the compact complement topology on $mathbb{R}$ to be $$tau'={Asubseteq Bbb{R}:A^Ctext{ is compact in }(Bbb{R},tau)} cup {emptyset }.$$
Hausdorff : let $G_1$ and $G_2$ be disjoint open sets containing $x$ and $y$ for $x ne y$ then
$$G_1 cap G_2 = varnothing $$
$$G_1^C cup G_2^C = Bbb R$$ but $Bbb R$ is not compact. So space is not Hausdorff.
Connectedness: Will the above also work for connectedness? Showing it's connected?
And are there any easy proof on compactness? I already saw this link but can't understand.
general-topology compactness connectedness
edited Feb 1 at 17:07
Exp ikx
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
$endgroup$
add a comment |
$begingroup$
For given the usual topology
$tau$ on $Bbb{R}$, define the compact complement topology on $mathbb{R}$ to be $$tau'={Asubseteq Bbb{R}:A^Ctext{ is compact in }(Bbb{R},tau)} cup {emptyset }.$$
Hausdorff : let $G_1$ and $G_2$ be disjoint open sets containing $x$ and $y$ for $x ne y$ then
$$G_1 cap G_2 = varnothing $$
$$G_1^C cup G_2^C = Bbb R$$ but $Bbb R$ is not compact. So space is not Hausdorff.
Connectedness: Will the above also work for connectedness? Showing it's connected?
And are there any easy proof on compactness? I already saw this link but can't understand.
general-topology compactness connectedness
edited Feb 1 at 17:07
Exp ikx
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
$endgroup$
add a comment |
$begingroup$
For given the usual topology
$tau$ on $Bbb{R}$, define the compact complement topology on $mathbb{R}$ to be $$tau'={Asubseteq Bbb{R}:A^Ctext{ is compact in }(Bbb{R},tau)} cup {emptyset }.$$
Hausdorff : let $G_1$ and $G_2$ be disjoint open sets containing $x$ and $y$ for $x ne y$ then
$$G_1 cap G_2 = varnothing $$
$$G_1^C cup G_2^C = Bbb R$$ but $Bbb R$ is not compact. So space is not Hausdorff.
Connectedness: Will the above also work for connectedness? Showing it's connected?
And are there any easy proof on compactness? I already saw this link but can't understand.
general-topology compactness connectedness
edited Feb 1 at 17:07
Exp ikx
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
$endgroup$
For given the usual topology
$tau$ on $Bbb{R}$, define the compact complement topology on $mathbb{R}$ to be $$tau'={Asubseteq Bbb{R}:A^Ctext{ is compact in }(Bbb{R},tau)} cup {emptyset }.$$
Hausdorff : let $G_1$ and $G_2$ be disjoint open sets containing $x$ and $y$ for $x ne y$ then
$$G_1 cap G_2 = varnothing $$
$$G_1^C cup G_2^C = Bbb R$$ but $Bbb R$ is not compact. So space is not Hausdorff.
Connectedness: Will the above also work for connectedness? Showing it's connected?
And are there any easy proof on compactness? I already saw this link but can't understand.
general-topology compactness connectedness
general-topology compactness connectedness
edited Feb 1 at 17:07
Exp ikx
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
edited Feb 1 at 17:07
Exp ikx
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
edited Feb 1 at 17:07
Exp ikx
4389
edited Feb 1 at 17:07
Exp ikx
4389
edited Feb 1 at 17:07
Exp ikx
4389
4389
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
asked Feb 1 at 16:37
Pranita GuptaPranita Gupta
123112
123112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
Compactness is not too hard: suppose ${U_i: i in I}$ is an open cover of $mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F subseteq Isetminus {i_0}$ that covers $U_{i_0}^c$ and then ${U_i: i in F cup {i_0}}$ is a finite subcover of our original cover. Hence the space is compact.
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
|
show 1 more comment
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1 Answer
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votes
1 Answer
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votes
$begingroup$
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
Compactness is not too hard: suppose ${U_i: i in I}$ is an open cover of $mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F subseteq Isetminus {i_0}$ that covers $U_{i_0}^c$ and then ${U_i: i in F cup {i_0}}$ is a finite subcover of our original cover. Hence the space is compact.
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
|
show 1 more comment
$begingroup$
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
Compactness is not too hard: suppose ${U_i: i in I}$ is an open cover of $mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F subseteq Isetminus {i_0}$ that covers $U_{i_0}^c$ and then ${U_i: i in F cup {i_0}}$ is a finite subcover of our original cover. Hence the space is compact.
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
|
show 1 more comment
$begingroup$
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
Compactness is not too hard: suppose ${U_i: i in I}$ is an open cover of $mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F subseteq Isetminus {i_0}$ that covers $U_{i_0}^c$ and then ${U_i: i in F cup {i_0}}$ is a finite subcover of our original cover. Hence the space is compact.
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
Compactness is not too hard: suppose ${U_i: i in I}$ is an open cover of $mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F subseteq Isetminus {i_0}$ that covers $U_{i_0}^c$ and then ${U_i: i in F cup {i_0}}$ is a finite subcover of our original cover. Hence the space is compact.
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 17:43
Henno BrandsmaHenno Brandsma
116k349127
116k349127
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
|
show 1 more comment
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity.
$endgroup$
– Pranita Gupta
Feb 1 at 17:49
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
@PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover.
$endgroup$
– Henno Brandsma
Feb 1 at 17:52
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
$begingroup$
I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite .
$endgroup$
– Pranita Gupta
Feb 1 at 17:57
1
1
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
$begingroup$
@PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did.
$endgroup$
– Henno Brandsma
Feb 1 at 17:59
1
1
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
$begingroup$
@PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover.
$endgroup$
– Henno Brandsma
Feb 1 at 18:05
|
show 1 more comment
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