# Small uncountable cardinals related to $\sigma$-continuity

A function $$f:X\to Y$$ is defined to be $$\sigma$$-continuous (resp. $$\bar \sigma$$-continuous) if there exists a countable (closed) cover $$\mathcal C$$ of $$X$$ such that the restriction $$f{\restriction}C$$ is continuous for every $$C\in\mathcal C$$.

Theorem 7.0 and Lemmas 7.4, 7.5 from Todorcevic's book "Partition Problems in Topology" imply the following

Theorem. Under MA every function $$f:X\to \mathbb R$$ defined on a subset $$X\subset\mathbb R$$ of cardinality $$|X|<\mathfrak c$$ is $$\bar\sigma$$-continuous.

This theorem suggests introducing two (new?) small uncountable cardinals $$\sigma$$ and $$\bar\sigma$$.

By definition, $$\sigma$$ (resp. $$\bar \sigma$$) is the smallest cardinality $$|f|$$ of a function $$f\subset\mathbb R\times\mathbb R$$, which is not $$\sigma$$-continuous (resp. $$\bar\sigma$$-continuous).

It is clear that $$\omega_1\le\bar\sigma\le\sigma\le\mathfrak c$$ and $$\bar\sigma\le\mathfrak q_0$$, where $$\mathfrak q_0$$ is the smallest cardinality of a subset $$X\subset\mathbb R$$, which is not a $$Q$$-set. We recall that a subset $$X\subset\mathbb R$$ is a $$Q$$-set if each subset of $$X$$ is of type $$F_\sigma$$-in $$X$$. It is easy to show that $$\bar\sigma<\mathfrak q_0$$ implies that $$\bar\sigma=\sigma$$.

Problem 1. Is $$\bar \sigma=\sigma$$ in ZFC?

Problem 2. Is $$\mathfrak p\le \bar\sigma$$?

Problem 3. Is $$\sigma$$ or $$\bar\sigma$$ equal to some known small uncountable cardinals? For example, is $$\bar\sigma=\mathfrak q_0$$?

Problem 4. Is there any non-trivial upper bound for the cardinal $$\sigma$$?

Problem 5. Let $$X$$ be a $$Q$$-set in $$\mathbb R$$. Is every function $$f:X\to\mathbb R$$ $$\bar\sigma$$-continuous? $$\sigma$$-continuous?

Remark 2. The answer to Problem 5 is affirmative if $$\mathfrak q_0=\mathfrak q$$, where $$\mathfrak q=\min\{\kappa:$$ no subset $$X\subset\mathbb R$$ of cardinality $$|X|\ge\kappa$$ is a $$Q$$-set in $$\mathbb R\}$$. It is known that $$\mathfrak q_0\le\mathfrak q\le \log(\mathfrak c^+)\le\mathfrak c$$ where $$\log(\mathfrak c^+)=\min\{\kappa:2^\kappa>\mathfrak c\}$$. The consistency of $$\mathfrak q_0<\mathfrak q$$ was proved by Judah and Shelah.

Added in Edit. Problem 5 has negative answer under the assumption $$\mathfrak q_0<\min\{\mathfrak b,\mathfrak q\}$$. Under this assumption we can find two subsets $$X,Y\subset\mathbb R$$ such that $$|X|=|Y|=\mathfrak q_0$$, $$X$$ is a $$Q$$-set, and $$Y$$ is not a $$Q$$-set. Using the strict inequality $$|X|<\mathfrak b$$, it can be shown that any bijective function $$f:X\to Y$$ is not $$\sigma$$-continuous. This also shows that $$\sigma=\bar\sigma=\mathfrak q_0$$ under $$\mathfrak q_0<\min\{\mathfrak b,\mathfrak q\}$$.

However this answer to Problem 5 has sense only if the following problem has an affirmative answer.

Problem 6. Is the strict inequality $$\mathfrak q_0<\min\{\mathfrak b,\mathfrak q\}$$ consistent?

• Might I suggest using a different letter, if this continues outside of this question? If nothing else, $\sigma$ is well-overused in mathematics, and usually somehow indicates $\omega_1$. So perhaps other letters can also have their time in the sun. – Asaf Karagila Mar 5 at 11:30
• @AsafKaragila Please suggest a different letter. In fact, I thought that these cardinals $\sigma$ and $\bar\sigma$ are equal to some known cardinals, so there will be no necessity to introduce new notations. And for the cardinal $\bar \sigma$ exactly this happened: it appeared to be equal to the known cardinal $\mathfrak q_0$. Maybe for $\sigma$ the situation is the same? Of course $\sigma$ is just a temporary symbol which should not be fixed for this small uncountable cardinal. It is like $\mathfrak{ridiculous}$ and $\mathfrak{stupid}$, if you remember such two temporary notations. – Taras Banakh Mar 5 at 11:48
• @AsafKaragila In fact, I would like to collect known information on the cardinal $\sigma$ and then decide if it deserve a special notation. In the latter case some reasonable notation (better than $\sigma$) will be chosen. – Taras Banakh Mar 5 at 11:51

Forcing to add uncountably many Cohen reals makes $$\sigma = \aleph_1$$.

To see this, suppose $$X$$ is an uncountable set of mutually generic Cohen reals over $$V$$. To be precise: mutually generic'' means that if $$x \in X$$ and $$Y \subseteq X$$, then $$x \in V[Y]$$ if and only if $$x \in Y$$. We wish to construct a function $$f \in V[X]$$ from $$A$$ to $$\mathbb R$$, where $$A \subseteq \mathbb R$$ and $$|A| = \aleph_1$$, and then prove $$f$$ is not $$\sigma$$-continuous.

The construction is easy: let $$A,B \subseteq X$$ such that $$|A| = |B| = \aleph_1$$ and $$A \cap B = \emptyset$$, and let $$f$$ be any bijection $$A \rightarrow B$$.

To prove $$f$$ is not $$\sigma$$-continuous, we will actually prove something a bit stronger: if $$A'$$ is any uncountable subset of $$A$$, then $$f \restriction A'$$ is not continuous. This implies $$f$$ is not $$\sigma$$-continuous, because if $$\mathcal C$$ is any countable cover of $$A$$, then some member of $$\mathcal C$$ must be uncountable.

So suppose $$A' \subseteq A$$ is uncountable. Let $$Q$$ be a countable dense subset of $$A'$$. Note that $$f \restriction Q$$ is a hereditarily countable set (it's a countable set of pairs of reals). Because the poset for adding Cohen reals has the ccc, every hereditarily countable set in the extension has a hereditarily countable name in the ground model. A hereditarily countable name for $$f \restriction Q$$ name only mentions'' bits of the Cohen forcing corresponding to countably many members of $$X$$. (More precisely, if we are forcing over $$V$$ with $$\mathrm{Fn}(\kappa \times \omega,2)$$, then there is a countable $$Z \subseteq \kappa$$ such that the name in question is really a $$\mathrm{Fn}(Z \times \omega,2)$$-name. Importantly, this means that we can compute $$f \restriction Q$$ if we know only what the generic looks like on $$Z \times \omega$$.) Hence $$f \restriction Q \in V[C]$$ for some countable $$C \subseteq X$$.

Because $$A'$$ is uncountable, there is some $$a \in A'$$ such that neither $$a$$ nor $$b = f(a)$$ are in $$C$$. Recall that $$b \neq a$$ (because $$A \cap B = \emptyset$$). Now, if $$f \restriction A'$$ were continuous, then, because $$Q$$ is dense in $$A'$$, we could compute $$b$$ from $$a$$ and $$f \restriction Q$$. Thus if $$f \restriction A'$$ were continuous we would have $$b \in V[C \cup \{a\}]$$. But this is nonsense: our Cohen reals were mutually generic, and in particular $$b$$ is Cohen-generic over $$V[C \cup \{a\}]$$.

It seems that the equality $$\sigma=\aleph_1$$ established by Will Brian in the Cohen model, can be derived from the following upper bound for $$\sigma$$, which answers Problem 4.

Theorem 1. $$\sigma\le\min\{\mathrm{non}(\mathcal N),\mathrm{non}(\mathcal M)\}$$.

Proof. Following Zakrzewski, we define a subset $$A$$ of a perfect Polish space $$X$$ to be universally meager if for any injective continuous map $$f:P\to X$$ of a perfect Polish space $$P$$ the preimage $$f^{-1}(A)$$ in meager in $$P$$. By a theorem of Grzegorek, the real line $$X$$ contains a universall meager subset $$A$$ of cardinality $$|A|=\mathrm{non}(\mathcal M)$$, where $$\mathrm{non}(\mathcal M)$$ stands for the smallest cardinality of a non-meager subset of the real line. Take any bijective map $$f:B\to A$$ from a non-meager set $$B$$ of $$\mathbb R$$ onto a universally meager set $$A\subset\mathbb R$$ and observe that it is not $$\sigma$$-continuous. Then $$\sigma\le|f|=\mathrm{non}(\mathcal M)$$.

A subet $$A\subset\mathbb R$$ has universal measure zero if $$\mu(A)=0$$ for any Borel continuous probability measure $$\mu$$ on $$\mathbb R$$. By a result of Grzegorek cited by Miller in his Handbook survey, the real line contains a subset $$A$$ of universal measure zero that has cardinality $$|A|=\mathrm{non}(\mathcal L)$$ where $$\mathrm{non}(\mathcal L)$$ is the smallest cardinality $$|B|$$ of a subset $$B\subset \mathbb R$$ that does not belong to the $$\sigma$$-ideal $$\mathcal L$$ of sets of Lebesgue measure zero in the real line. Take any bijection $$g:B\to A$$ and observe that $$g$$ is not $$\sigma$$-continuous, which implies that $$\sigma\le|g|=\mathrm{non}(\mathcal L)$$. $$\square$$

Problem 2 and the second part of Problem 3 have affirmative answers because of the following

Theorem 2. $$\mathfrak p\le \bar\sigma=\mathfrak q_0\le\mathfrak b$$.

Proof. The proof of the inequalities $$\mathfrak p\le\mathfrak q_0\le\mathfrak b$$ can be found in this paper of Banakh, Machura and Zdomskyy.

To see that $$\mathfrak q_0\le\bar\sigma$$, observe that for any subset $$X\subset\mathbb R$$ of cardinality $$|X|<\bar\sigma$$ and any subset $$A\subset X$$ the characteristic function $$\chi_A:X\to \{0,1\}$$ of $$A$$ is $$\bar\sigma$$-continuous, which implies that $$A$$ is of type $$F_\sigma$$ in $$X$$.

To see that $$\bar\sigma\le\mathfrak q_0$$, take any function $$f:X\to [0,1]$$ on a set $$X\subset[0,1]$$ of cardinality $$|X|<\mathbb q_0$$. Let $$Y=f(X)$$ and consider the graph $$\Gamma=\{(x,f(x)):x\in X\}$$ of $$f$$ in $$X\times Y$$. Since $$|X\times Y|<\mathfrak q_0$$, the graph $$\Gamma$$ is an $$F_\sigma$$-set in $$X\times Y$$. Then $$\Gamma=\bigcup_{n\in\omega}\Gamma_n$$ for some closed subsets in $$X\times Y$$.

For every $$n\in\omega$$, consider the closure $$\bar \Gamma_n$$ of $$\Gamma_n$$ in $$[0,1]\times[0,1]$$ and the projection $$pr:\bar \Gamma_n\to[0,1]$$, $$pr:(x,y)\mapsto x$$. The compactness of $$\bar\Gamma_n$$ implies that $$D_n:=pr(\Gamma_n)$$ is compact and the set $$M_n=\{y\in D_n:|pr^{-1}(y)|>1\}$$ is $$\sigma$$-compact and disjoint with $$X\cap D_n$$. Since $$|X\cap D_n|<\mathfrak q_0\le\mathfrak b$$, the set $$X\cap D_n$$ is contained in the countable union $$\bigcup_{m\in\omega}K_{n,m}$$ of compact subsets of the Polish space $$D_n\setminus M_n$$. For every $$m\in\omega$$ the set $$\bar \Gamma_n\cap (K_{n,m}\times[0,1])$$ is compact and coincides with the graph of a (continuous) function. This implies that the restriction $$f{\restriction}X\cap K_{n,m}$$ is continuous. Since $$X=\bigcup_{n,m\in\omega}K_{n,m}$$, the function $$f$$ is $$\bar\sigma$$-continuous. $$\square$$

Theorem 1 suggests the following

Problem I. Is $$\sigma\le\mathrm{non}(\mathcal I)$$ for any ccc $$\sigma$$-ideal $$\mathcal I$$ with Borel base on a perfect Polish space?

Problem E. Is $$\sigma\le \mathrm{non}({\mathcal E})$$ for the $$\sigma$$-ideal $$\mathcal E$$ generated by closed Lebesgue null subsets of the real line?