What are the basis for co-countable and co-finite topologies
$begingroup$
I have just learned co-countable and co-finite topologies
But it was never mentioned what are the bases for these topologies and I cannot seem to find any reference to it anywhere.
For example, in the standard topology $tau_{standard}$ on $Bbb{R}$, a basis is $mathcal{B} = {(a,b) subset Bbb{R} | a, b in Bbb{R}}$
Is there something similar for those topologies?
general-topology
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
$endgroup$
add a comment |
$begingroup$
I have just learned co-countable and co-finite topologies
But it was never mentioned what are the bases for these topologies and I cannot seem to find any reference to it anywhere.
For example, in the standard topology $tau_{standard}$ on $Bbb{R}$, a basis is $mathcal{B} = {(a,b) subset Bbb{R} | a, b in Bbb{R}}$
Is there something similar for those topologies?
general-topology
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
$endgroup$
$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
$endgroup$
– Shamisen Expert
Jun 9 '16 at 15:01
$begingroup$
True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
$endgroup$
– André Nicolas
Jun 9 '16 at 15:07
add a comment |
$begingroup$
I have just learned co-countable and co-finite topologies
But it was never mentioned what are the bases for these topologies and I cannot seem to find any reference to it anywhere.
For example, in the standard topology $tau_{standard}$ on $Bbb{R}$, a basis is $mathcal{B} = {(a,b) subset Bbb{R} | a, b in Bbb{R}}$
Is there something similar for those topologies?
general-topology
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
$endgroup$
I have just learned co-countable and co-finite topologies
But it was never mentioned what are the bases for these topologies and I cannot seem to find any reference to it anywhere.
For example, in the standard topology $tau_{standard}$ on $Bbb{R}$, a basis is $mathcal{B} = {(a,b) subset Bbb{R} | a, b in Bbb{R}}$
Is there something similar for those topologies?
general-topology
general-topology
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
edited Jun 9 '16 at 18:18
Shamisen Expert
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
asked Jun 9 '16 at 14:38
Shamisen ExpertShamisen Expert
2,77721945
2,77721945
$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
$endgroup$
– Shamisen Expert
Jun 9 '16 at 15:01
$begingroup$
True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
$endgroup$
– André Nicolas
Jun 9 '16 at 15:07
add a comment |
$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
$endgroup$
– Shamisen Expert
Jun 9 '16 at 15:01
$begingroup$
True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
$endgroup$
– André Nicolas
Jun 9 '16 at 15:07
$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
$endgroup$
– Shamisen Expert
Jun 9 '16 at 15:01
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
$endgroup$
– Shamisen Expert
Jun 9 '16 at 15:01
$begingroup$
True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
$endgroup$
– André Nicolas
Jun 9 '16 at 15:07
$begingroup$
True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
$endgroup$
– André Nicolas
Jun 9 '16 at 15:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A basis is just one tool for describing a topology. Most topologies have many different bases that generate them, but most of these bases aren't useful. The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis. After all, what could be easier to think about than simply saying a set has finite (or countable) complement? The open sets themselves already have a very simple structure. Moreover, it's not as though open sets are built out of small "balls" around each point. For instance, in the cofinite topology, if $U$ is open and $xin U$ and you take an open set $V$ such that $xin Vsubseteq U$, then $V$ must already contain all but finitely many points of $U$. So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $mathcal{B}={Usubset X:|Xsetminus U|=1}$ of complements of single points. Every cofinite set is a finite intersection of these sets, so they do form a subbasis. In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $mathcal{B}$, or a map $f:Yto X$ is continuous iff $f^{-1}(U)$ is open for each $Uinmathcal{B}$.
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
add a comment |
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$begingroup$
A basis is just one tool for describing a topology. Most topologies have many different bases that generate them, but most of these bases aren't useful. The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis. After all, what could be easier to think about than simply saying a set has finite (or countable) complement? The open sets themselves already have a very simple structure. Moreover, it's not as though open sets are built out of small "balls" around each point. For instance, in the cofinite topology, if $U$ is open and $xin U$ and you take an open set $V$ such that $xin Vsubseteq U$, then $V$ must already contain all but finitely many points of $U$. So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $mathcal{B}={Usubset X:|Xsetminus U|=1}$ of complements of single points. Every cofinite set is a finite intersection of these sets, so they do form a subbasis. In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $mathcal{B}$, or a map $f:Yto X$ is continuous iff $f^{-1}(U)$ is open for each $Uinmathcal{B}$.
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
add a comment |
$begingroup$
A basis is just one tool for describing a topology. Most topologies have many different bases that generate them, but most of these bases aren't useful. The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis. After all, what could be easier to think about than simply saying a set has finite (or countable) complement? The open sets themselves already have a very simple structure. Moreover, it's not as though open sets are built out of small "balls" around each point. For instance, in the cofinite topology, if $U$ is open and $xin U$ and you take an open set $V$ such that $xin Vsubseteq U$, then $V$ must already contain all but finitely many points of $U$. So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $mathcal{B}={Usubset X:|Xsetminus U|=1}$ of complements of single points. Every cofinite set is a finite intersection of these sets, so they do form a subbasis. In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $mathcal{B}$, or a map $f:Yto X$ is continuous iff $f^{-1}(U)$ is open for each $Uinmathcal{B}$.
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
add a comment |
$begingroup$
A basis is just one tool for describing a topology. Most topologies have many different bases that generate them, but most of these bases aren't useful. The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis. After all, what could be easier to think about than simply saying a set has finite (or countable) complement? The open sets themselves already have a very simple structure. Moreover, it's not as though open sets are built out of small "balls" around each point. For instance, in the cofinite topology, if $U$ is open and $xin U$ and you take an open set $V$ such that $xin Vsubseteq U$, then $V$ must already contain all but finitely many points of $U$. So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $mathcal{B}={Usubset X:|Xsetminus U|=1}$ of complements of single points. Every cofinite set is a finite intersection of these sets, so they do form a subbasis. In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $mathcal{B}$, or a map $f:Yto X$ is continuous iff $f^{-1}(U)$ is open for each $Uinmathcal{B}$.
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
$endgroup$
A basis is just one tool for describing a topology. Most topologies have many different bases that generate them, but most of these bases aren't useful. The point of a basis is that it should ideally consist of sets that are easier to think about than general open sets.
In the case of the cofinite or cocountable topology, there is no such easier basis. After all, what could be easier to think about than simply saying a set has finite (or countable) complement? The open sets themselves already have a very simple structure. Moreover, it's not as though open sets are built out of small "balls" around each point. For instance, in the cofinite topology, if $U$ is open and $xin U$ and you take an open set $V$ such that $xin Vsubseteq U$, then $V$ must already contain all but finitely many points of $U$. So $V$ is not really going to be any simpler to think about than $U$ itself.
For the cofinite topology, however, there is a natural subbasis that is somewhat easier to think about and sometimes useful: namely, the collection $mathcal{B}={Usubset X:|Xsetminus U|=1}$ of complements of single points. Every cofinite set is a finite intersection of these sets, so they do form a subbasis. In particular, this is useful (for instance) because it tells you that a topology contains the cofinite topology iff it contains each element of $mathcal{B}$, or a map $f:Yto X$ is continuous iff $f^{-1}(U)$ is open for each $Uinmathcal{B}$.
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
answered Jun 9 '16 at 18:25
Eric WofseyEric Wofsey
187k14216344
187k14216344
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
add a comment |
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
Interesting I learned something neu today.
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:42
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
$begingroup$
But the proof of that set is a subbasis is probably really hard, the collection looks unfriendly
$endgroup$
– Shamisen Expert
Jun 9 '16 at 18:48
1
1
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
No, it's very simple. If $Vsubseteq X$ is any set, then $V$ is just the intersection of all the sets $Xsetminus{x}$ where $x$ ranges over all the elements of $Xsetminus V$. So if $Xsetminus V$ is finite, then $V$ is a finite intersection of sets in $mathcal{B}$.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 18:56
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Okay I see, so $V = bigcaplimits_{{x} in X backslash V} X backslash {x}$, where $X backslash {x} in mathcal{B}$ correct? $X backslash {x} in mathcal{B}$ since $X backslash (Xbackslash {x}) = {x}$ is a one point set
$endgroup$
– Shamisen Expert
Jun 9 '16 at 19:02
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
$begingroup$
Yes, that's exactly right.
$endgroup$
– Eric Wofsey
Jun 9 '16 at 19:54
add a comment |
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$begingroup$
We can even explicitly describe all open sets. If $X$ is the underlying set, for co-countable the opens are the empty set and the subsets of $X$ whose complement is countable.
$endgroup$
– André Nicolas
Jun 9 '16 at 14:56
$begingroup$
@AndréNicolas Can you clarify what you mean by "explicitly desribe all open sets". Are you implying that we only specify basis in the case when we cannot explicitly describe all open sets? But we can do that even for $tau_{standard}$ because all the open sets are just set of the form $(a,b), varnothing$ and $mathbb{X}$, their unions and finite intersections
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– Shamisen Expert
Jun 9 '16 at 15:01
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True, one can specify the open sets in the usual topology, though the specification is certainly more complicated than for the co-countable or co-finite topology on a given set. In the case of the topologies you asked about, the reasonable basis to take is all open sets, or all non-empty open.
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– André Nicolas
Jun 9 '16 at 15:07