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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?

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  • 2
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Mar 5 at 11:30
  • $\begingroup$ @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. $\endgroup$ – Taras Banakh Mar 5 at 11:48
  • $\begingroup$ @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. $\endgroup$ – Taras Banakh Mar 5 at 11:51
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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 $V[C \cup \{a\}]$.

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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?

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