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Noah Schweber
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This should be a comment, but it's too long - here's a $\mathsf{ZFC}+\mathsf{CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $\operatorname{Space}(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC}+\mathsf{CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $\operatorname{Space}(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $\operatorname{Space}(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)

(At a glance, making each $\{X: r\in X\}$ clopen won't affect this analysis. This would result in a Hausdorff space, which would be nice; $\operatorname{Space}(\mathfrak{F})$ itself isn't even $T_1$.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

This should be a comment, but it's too long - here's a $\mathsf{ZFC}+\mathsf{CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $\operatorname{Space}(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC}+\mathsf{CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $\operatorname{Space}(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $\operatorname{Space}(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

This should be a comment, but it's too long - here's a $\mathsf{ZFC}+\mathsf{CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $\operatorname{Space}(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC}+\mathsf{CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $\operatorname{Space}(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $\operatorname{Space}(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)

(At a glance, making each $\{X: r\in X\}$ clopen won't affect this analysis. This would result in a Hausdorff space, which would be nice; $\operatorname{Space}(\mathfrak{F})$ itself isn't even $T_1$.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

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LSpice
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This should be a comment, but it's too long - here's a $\mathsf{ZFC+CH}$$\mathsf{ZFC}+\mathsf{CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $Space(\mathfrak{F})$$\operatorname{Space}(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC+CH}$$\mathsf{ZFC}+\mathsf{CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $Space(\mathfrak{F})$$\operatorname{Space}(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $Space(\mathfrak{F})$$\operatorname{Space}(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

This should be a comment, but it's too long - here's a $\mathsf{ZFC+CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $Space(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC+CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $Space(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $Space(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

This should be a comment, but it's too long - here's a $\mathsf{ZFC}+\mathsf{CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $\operatorname{Space}(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC}+\mathsf{CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $\operatorname{Space}(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $\operatorname{Space}(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).

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Noah Schweber
  • 20.5k
  • 10
  • 110
  • 331

This should be a comment, but it's too long - here's a $\mathsf{ZFC+CH}$ example:

Let $\mathfrak{F}$ be the set of full-measure subsets of $\mathbb{R}$ (we could also take the set of non-meager subsets, or so on), and let $Space(\mathfrak{F})$ be the topological space whose points are the full-measure subsets of $\mathbb{R}$ and whose topology comes from the"canonical generator" $$\rho:r\mapsto\{X\in\mathfrak{F}: r\in X\}.$$ For this space instances of covering correspond to non-null sets: we have $\rho(g)\subseteq\bigcup_{f\in U}\rho(f)$ iff $g\in U$ or $U$ is non-null.

A standard transfinite recursion argument shows in $\mathsf{ZFC+CH}$ that for every size-continuum set of non-null sets $(N_i)_{i\in\mathbb{R}}$ there is a non-null set $B$ such that $B\not\supseteq N_i$ for any $i$. This implies a weak form of non-quickness for $Space(\mathfrak{F})$, namely that the canonical generator $\rho$ does not witness quickness, and it's not hard to extend this to arbitrary generators. So $Space(\mathfrak{F})$ is not quick.

(Annoyingly I don't see that if a space is quick then all generators should witness that; luckily, that isn't an issue here.)


Of course this example does not work under determinacy, since every non-null set has a non-null $F_\sigma$ subset and there are only continuum-many of those. But I suspect a more complicated ideal of sets than the null (or meager, or etc.) ideal will do the job. Note that given an $I$-indexed subbase for a $T_0$ topological space $\mathcal{X}$ there is a canonical homeomorphic copy of $\mathcal{X}$ whose points are subsets of $I$, so this isn't really a significant shift.

At the same time, the above raises a separate question:

What can we say, in $\mathsf{ZFC}$, about the minimal cardinality of a set of non-null sets $\mathfrak{A}$ such that every non-null set contains an element of $\mathfrak{A}$? (Or non-meager, or so on.)

I don't really have anything to say on this point, but I'd be interested in information about it (and may ask a separate question about it later on).