A property between “separable” and “second countable”Second-countable implies separable/Axiom...

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A property between "separable" and "second countable"



A property between “separable” and “second countable”


Second-countable implies separable/Axiom countable choiceDoes Seperable + First Countable + Sigma-Locally Finite Basis Imply Second Countable?separable space and open coversSecond Countable, First Countable, and Separable SpacesProve that $l_2$ is a second countable space and is separable.If a topological space $X$ is $E_{2}$ (second-countable), then it's separable.X is second countable then $Asubset X$ has an accumulation pointIf a topological space can be partitioned into finitely many second countable subspaces, is it second countable? [Collecting examples]First countable separable spaces are not second countable.






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







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$begingroup$


Let $(X, tau)$ be a topological space. It is second countable if it has a countable basis $B subseteq tau$. It is separable if there exists a countable $S subseteq X$ such that $O cap S neq emptyset$ for every $O in tau$. It is well known that second countability is strictly stronger than separability.



I'm working on something hinges on an intermediate property: "there exists a countable subset $C subseteq tau$ [edit: with each $C$-member nonempty!] that is dense in $tau$, in the sense that for all $O in tau$, there exists $P in C$ such that $P subseteq O$."



Is there a common name for this property? I will call it "property C" for now.



Second countability implies property C (since a countable basis for $tau$ is dense in $tau$), which implies separability (choose one member from each $P in C$ and the set of all the choices serves as the $S$ in the definition of separability). The Moore plane is an example of a topology that has property C but is not second countable.



Are there examples of topological spaces that are separable but do not have property C?










share|cite|improve this question











$endgroup$














  • $begingroup$
    I think that the property C implies first countable.
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    @MarcoLecci that is quite false..
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    @HennoBrandsma yes you right
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    Note that a separable and first countable space will also have your property C.
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
    $endgroup$
    – Henno Brandsma
    7 hours ago


















6












$begingroup$


Let $(X, tau)$ be a topological space. It is second countable if it has a countable basis $B subseteq tau$. It is separable if there exists a countable $S subseteq X$ such that $O cap S neq emptyset$ for every $O in tau$. It is well known that second countability is strictly stronger than separability.



I'm working on something hinges on an intermediate property: "there exists a countable subset $C subseteq tau$ [edit: with each $C$-member nonempty!] that is dense in $tau$, in the sense that for all $O in tau$, there exists $P in C$ such that $P subseteq O$."



Is there a common name for this property? I will call it "property C" for now.



Second countability implies property C (since a countable basis for $tau$ is dense in $tau$), which implies separability (choose one member from each $P in C$ and the set of all the choices serves as the $S$ in the definition of separability). The Moore plane is an example of a topology that has property C but is not second countable.



Are there examples of topological spaces that are separable but do not have property C?










share|cite|improve this question











$endgroup$














  • $begingroup$
    I think that the property C implies first countable.
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    @MarcoLecci that is quite false..
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    @HennoBrandsma yes you right
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    Note that a separable and first countable space will also have your property C.
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
    $endgroup$
    – Henno Brandsma
    7 hours ago














6












6








6


1



$begingroup$


Let $(X, tau)$ be a topological space. It is second countable if it has a countable basis $B subseteq tau$. It is separable if there exists a countable $S subseteq X$ such that $O cap S neq emptyset$ for every $O in tau$. It is well known that second countability is strictly stronger than separability.



I'm working on something hinges on an intermediate property: "there exists a countable subset $C subseteq tau$ [edit: with each $C$-member nonempty!] that is dense in $tau$, in the sense that for all $O in tau$, there exists $P in C$ such that $P subseteq O$."



Is there a common name for this property? I will call it "property C" for now.



Second countability implies property C (since a countable basis for $tau$ is dense in $tau$), which implies separability (choose one member from each $P in C$ and the set of all the choices serves as the $S$ in the definition of separability). The Moore plane is an example of a topology that has property C but is not second countable.



Are there examples of topological spaces that are separable but do not have property C?










share|cite|improve this question











$endgroup$




Let $(X, tau)$ be a topological space. It is second countable if it has a countable basis $B subseteq tau$. It is separable if there exists a countable $S subseteq X$ such that $O cap S neq emptyset$ for every $O in tau$. It is well known that second countability is strictly stronger than separability.



I'm working on something hinges on an intermediate property: "there exists a countable subset $C subseteq tau$ [edit: with each $C$-member nonempty!] that is dense in $tau$, in the sense that for all $O in tau$, there exists $P in C$ such that $P subseteq O$."



Is there a common name for this property? I will call it "property C" for now.



Second countability implies property C (since a countable basis for $tau$ is dense in $tau$), which implies separability (choose one member from each $P in C$ and the set of all the choices serves as the $S$ in the definition of separability). The Moore plane is an example of a topology that has property C but is not second countable.



Are there examples of topological spaces that are separable but do not have property C?







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago







Doug McLellan

















asked 8 hours ago









Doug McLellanDoug McLellan

4312 silver badges8 bronze badges




4312 silver badges8 bronze badges















  • $begingroup$
    I think that the property C implies first countable.
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    @MarcoLecci that is quite false..
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    @HennoBrandsma yes you right
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    Note that a separable and first countable space will also have your property C.
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
    $endgroup$
    – Henno Brandsma
    7 hours ago


















  • $begingroup$
    I think that the property C implies first countable.
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    @MarcoLecci that is quite false..
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    @HennoBrandsma yes you right
    $endgroup$
    – Marco Lecci
    8 hours ago






  • 1




    $begingroup$
    Note that a separable and first countable space will also have your property C.
    $endgroup$
    – Henno Brandsma
    8 hours ago










  • $begingroup$
    Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
    $endgroup$
    – Henno Brandsma
    7 hours ago
















$begingroup$
I think that the property C implies first countable.
$endgroup$
– Marco Lecci
8 hours ago




$begingroup$
I think that the property C implies first countable.
$endgroup$
– Marco Lecci
8 hours ago




1




1




$begingroup$
@MarcoLecci that is quite false..
$endgroup$
– Henno Brandsma
8 hours ago




$begingroup$
@MarcoLecci that is quite false..
$endgroup$
– Henno Brandsma
8 hours ago












$begingroup$
@HennoBrandsma yes you right
$endgroup$
– Marco Lecci
8 hours ago




$begingroup$
@HennoBrandsma yes you right
$endgroup$
– Marco Lecci
8 hours ago




1




1




$begingroup$
Note that a separable and first countable space will also have your property C.
$endgroup$
– Henno Brandsma
8 hours ago




$begingroup$
Note that a separable and first countable space will also have your property C.
$endgroup$
– Henno Brandsma
8 hours ago












$begingroup$
Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
$endgroup$
– Henno Brandsma
7 hours ago




$begingroup$
Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course.
$endgroup$
– Henno Brandsma
7 hours ago










2 Answers
2






active

oldest

votes


















2














$begingroup$

Consider $mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $mathbb R$ is dense since an open subset of $mathbb R$ can only miss finitely many points of $A$.



Let $mathcal B$ be a countable family of non-empty open subsets of $mathbb R$. Then $bigcap_{B in mathcal B} B$ misses countably many points of $mathbb R$ at most. It follows that some $x in mathbb R$ lies in every element of $mathcal B$. Thus, the open subset $mathbb R - {x}$ does not contain any element of $mathcal B$.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
    $endgroup$
    – Doug McLellan
    8 hours ago










  • $begingroup$
    @DougMcLellan happy to help! And yes, you're right. Fixed now.
    $endgroup$
    – Ayman Hourieh
    8 hours ago










  • $begingroup$
    sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
    $endgroup$
    – Doug McLellan
    7 hours ago






  • 1




    $begingroup$
    In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
    $endgroup$
    – Ayman Hourieh
    7 hours ago










  • $begingroup$
    [you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
    $endgroup$
    – Doug McLellan
    7 hours ago



















4














$begingroup$

What you're looking for is the concept of a $pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $mathcal{P}$ such that any non-empty open
subset of $X$ contains a member of $mathcal{P}$. (The collection is downward-dense in the poset $(mathcal{T}setminus{emptyset}, subseteq)$ is another way of putting it)



The minimal size of a $pi$-base for $X$ is denoted $pi w(X)$ (rounded up to $aleph_0$ if necessary, in Juhasz it's $pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $pi$-weight or $pi w(X)=aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $pi$-weight, or having a countable $pi$-base as a name, being a bit more descriptive.



As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{Bbb R}$ in the product topology is separable but has no countable $pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $pi$-base are all equivalent.



There also exists the notion of a local $pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $pichi(x,X)$ for the minimal size of such a collection, etc.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
    $endgroup$
    – Doug McLellan
    1 hour ago










  • $begingroup$
    @DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
    $endgroup$
    – Henno Brandsma
    1 hour ago














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














$begingroup$

Consider $mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $mathbb R$ is dense since an open subset of $mathbb R$ can only miss finitely many points of $A$.



Let $mathcal B$ be a countable family of non-empty open subsets of $mathbb R$. Then $bigcap_{B in mathcal B} B$ misses countably many points of $mathbb R$ at most. It follows that some $x in mathbb R$ lies in every element of $mathcal B$. Thus, the open subset $mathbb R - {x}$ does not contain any element of $mathcal B$.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
    $endgroup$
    – Doug McLellan
    8 hours ago










  • $begingroup$
    @DougMcLellan happy to help! And yes, you're right. Fixed now.
    $endgroup$
    – Ayman Hourieh
    8 hours ago










  • $begingroup$
    sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
    $endgroup$
    – Doug McLellan
    7 hours ago






  • 1




    $begingroup$
    In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
    $endgroup$
    – Ayman Hourieh
    7 hours ago










  • $begingroup$
    [you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
    $endgroup$
    – Doug McLellan
    7 hours ago
















2














$begingroup$

Consider $mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $mathbb R$ is dense since an open subset of $mathbb R$ can only miss finitely many points of $A$.



Let $mathcal B$ be a countable family of non-empty open subsets of $mathbb R$. Then $bigcap_{B in mathcal B} B$ misses countably many points of $mathbb R$ at most. It follows that some $x in mathbb R$ lies in every element of $mathcal B$. Thus, the open subset $mathbb R - {x}$ does not contain any element of $mathcal B$.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
    $endgroup$
    – Doug McLellan
    8 hours ago










  • $begingroup$
    @DougMcLellan happy to help! And yes, you're right. Fixed now.
    $endgroup$
    – Ayman Hourieh
    8 hours ago










  • $begingroup$
    sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
    $endgroup$
    – Doug McLellan
    7 hours ago






  • 1




    $begingroup$
    In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
    $endgroup$
    – Ayman Hourieh
    7 hours ago










  • $begingroup$
    [you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
    $endgroup$
    – Doug McLellan
    7 hours ago














2














2










2







$begingroup$

Consider $mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $mathbb R$ is dense since an open subset of $mathbb R$ can only miss finitely many points of $A$.



Let $mathcal B$ be a countable family of non-empty open subsets of $mathbb R$. Then $bigcap_{B in mathcal B} B$ misses countably many points of $mathbb R$ at most. It follows that some $x in mathbb R$ lies in every element of $mathcal B$. Thus, the open subset $mathbb R - {x}$ does not contain any element of $mathcal B$.






share|cite|improve this answer











$endgroup$



Consider $mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $mathbb R$ is dense since an open subset of $mathbb R$ can only miss finitely many points of $A$.



Let $mathcal B$ be a countable family of non-empty open subsets of $mathbb R$. Then $bigcap_{B in mathcal B} B$ misses countably many points of $mathbb R$ at most. It follows that some $x in mathbb R$ lies in every element of $mathcal B$. Thus, the open subset $mathbb R - {x}$ does not contain any element of $mathcal B$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 7 hours ago

























answered 8 hours ago









Ayman HouriehAyman Hourieh

32.1k4 gold badges67 silver badges124 bronze badges




32.1k4 gold badges67 silver badges124 bronze badges















  • $begingroup$
    brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
    $endgroup$
    – Doug McLellan
    8 hours ago










  • $begingroup$
    @DougMcLellan happy to help! And yes, you're right. Fixed now.
    $endgroup$
    – Ayman Hourieh
    8 hours ago










  • $begingroup$
    sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
    $endgroup$
    – Doug McLellan
    7 hours ago






  • 1




    $begingroup$
    In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
    $endgroup$
    – Ayman Hourieh
    7 hours ago










  • $begingroup$
    [you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
    $endgroup$
    – Doug McLellan
    7 hours ago


















  • $begingroup$
    brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
    $endgroup$
    – Doug McLellan
    8 hours ago










  • $begingroup$
    @DougMcLellan happy to help! And yes, you're right. Fixed now.
    $endgroup$
    – Ayman Hourieh
    8 hours ago










  • $begingroup$
    sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
    $endgroup$
    – Doug McLellan
    7 hours ago






  • 1




    $begingroup$
    In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
    $endgroup$
    – Ayman Hourieh
    7 hours ago










  • $begingroup$
    [you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
    $endgroup$
    – Doug McLellan
    7 hours ago
















$begingroup$
brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
$endgroup$
– Doug McLellan
8 hours ago




$begingroup$
brilliant, thank you. (i think you meant $bigcap_{B in mathcal{B}} B$ rather than $bigcup_{B in mathcal{B}} B$)
$endgroup$
– Doug McLellan
8 hours ago












$begingroup$
@DougMcLellan happy to help! And yes, you're right. Fixed now.
$endgroup$
– Ayman Hourieh
8 hours ago




$begingroup$
@DougMcLellan happy to help! And yes, you're right. Fixed now.
$endgroup$
– Ayman Hourieh
8 hours ago












$begingroup$
sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
$endgroup$
– Doug McLellan
7 hours ago




$begingroup$
sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P in B$ contains $O$, whereas we need an $O$ that contains no $P in B$.
$endgroup$
– Doug McLellan
7 hours ago




1




1




$begingroup$
In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
$endgroup$
– Ayman Hourieh
7 hours ago




$begingroup$
In my example, $O = mathbb R - {x}$ doesn't contain any element of $mathcal B$, since every element of $mathcal B$ contains $x$. Am I missing something?
$endgroup$
– Ayman Hourieh
7 hours ago












$begingroup$
[you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
$endgroup$
– Doug McLellan
7 hours ago




$begingroup$
[you are actually correct & i'm going to log off until my caffeine level exits the red zone!]
$endgroup$
– Doug McLellan
7 hours ago













4














$begingroup$

What you're looking for is the concept of a $pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $mathcal{P}$ such that any non-empty open
subset of $X$ contains a member of $mathcal{P}$. (The collection is downward-dense in the poset $(mathcal{T}setminus{emptyset}, subseteq)$ is another way of putting it)



The minimal size of a $pi$-base for $X$ is denoted $pi w(X)$ (rounded up to $aleph_0$ if necessary, in Juhasz it's $pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $pi$-weight or $pi w(X)=aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $pi$-weight, or having a countable $pi$-base as a name, being a bit more descriptive.



As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{Bbb R}$ in the product topology is separable but has no countable $pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $pi$-base are all equivalent.



There also exists the notion of a local $pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $pichi(x,X)$ for the minimal size of such a collection, etc.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
    $endgroup$
    – Doug McLellan
    1 hour ago










  • $begingroup$
    @DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
    $endgroup$
    – Henno Brandsma
    1 hour ago
















4














$begingroup$

What you're looking for is the concept of a $pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $mathcal{P}$ such that any non-empty open
subset of $X$ contains a member of $mathcal{P}$. (The collection is downward-dense in the poset $(mathcal{T}setminus{emptyset}, subseteq)$ is another way of putting it)



The minimal size of a $pi$-base for $X$ is denoted $pi w(X)$ (rounded up to $aleph_0$ if necessary, in Juhasz it's $pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $pi$-weight or $pi w(X)=aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $pi$-weight, or having a countable $pi$-base as a name, being a bit more descriptive.



As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{Bbb R}$ in the product topology is separable but has no countable $pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $pi$-base are all equivalent.



There also exists the notion of a local $pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $pichi(x,X)$ for the minimal size of such a collection, etc.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
    $endgroup$
    – Doug McLellan
    1 hour ago










  • $begingroup$
    @DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
    $endgroup$
    – Henno Brandsma
    1 hour ago














4














4










4







$begingroup$

What you're looking for is the concept of a $pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $mathcal{P}$ such that any non-empty open
subset of $X$ contains a member of $mathcal{P}$. (The collection is downward-dense in the poset $(mathcal{T}setminus{emptyset}, subseteq)$ is another way of putting it)



The minimal size of a $pi$-base for $X$ is denoted $pi w(X)$ (rounded up to $aleph_0$ if necessary, in Juhasz it's $pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $pi$-weight or $pi w(X)=aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $pi$-weight, or having a countable $pi$-base as a name, being a bit more descriptive.



As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{Bbb R}$ in the product topology is separable but has no countable $pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $pi$-base are all equivalent.



There also exists the notion of a local $pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $pichi(x,X)$ for the minimal size of such a collection, etc.






share|cite|improve this answer











$endgroup$



What you're looking for is the concept of a $pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $mathcal{P}$ such that any non-empty open
subset of $X$ contains a member of $mathcal{P}$. (The collection is downward-dense in the poset $(mathcal{T}setminus{emptyset}, subseteq)$ is another way of putting it)



The minimal size of a $pi$-base for $X$ is denoted $pi w(X)$ (rounded up to $aleph_0$ if necessary, in Juhasz it's $pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $pi$-weight or $pi w(X)=aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $pi$-weight, or having a countable $pi$-base as a name, being a bit more descriptive.



As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{Bbb R}$ in the product topology is separable but has no countable $pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $pi$-base are all equivalent.



There also exists the notion of a local $pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $pichi(x,X)$ for the minimal size of such a collection, etc.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 8 hours ago









Henno BrandsmaHenno Brandsma

130k4 gold badges53 silver badges137 bronze badges




130k4 gold badges53 silver badges137 bronze badges















  • $begingroup$
    Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
    $endgroup$
    – Doug McLellan
    1 hour ago










  • $begingroup$
    @DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
    $endgroup$
    – Henno Brandsma
    1 hour ago


















  • $begingroup$
    Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
    $endgroup$
    – Doug McLellan
    1 hour ago










  • $begingroup$
    @DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
    $endgroup$
    – Henno Brandsma
    1 hour ago
















$begingroup$
Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
$endgroup$
– Doug McLellan
1 hour ago




$begingroup$
Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff.
$endgroup$
– Doug McLellan
1 hour ago












$begingroup$
@DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
$endgroup$
– Henno Brandsma
1 hour ago




$begingroup$
@DougMcLellan Maybe a nicer challenge is to find a space $X$ with $aleph_0=d(X) < pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group?
$endgroup$
– Henno Brandsma
1 hour ago



















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