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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?


Added in Edit (13.09.2019). The second part of Problem 3 has affirmative solution: $\bar\sigma=\mathfrak q_0$. Problem 5 has an affirmative answer under an additional assumption $|X|<\mathfrak b$ (which follows from $\mathfrak q\le\mathfrak b$). The proofs are written in my answer below.

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    $\begingroup$ 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. $\sigma\le\min\{\mathrm{non}(\mathcal N),\mathrm{non}(\mathcal M)\}$. Theorem 1 is proved in this paper. This theorem 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?

Now we give a partial answer to Problem 5.

Theorem U (proved in Ustka on 12.09.2019). Let $X,Y$ be separable metrizable spaces. If $|X|<\mathfrak b$, then each Borel function $f:X\to Y$ is $\bar\sigma$-continuous.

Proof. We lose no generality assuming that the space $Y$ is Polish and $X$ is a subspace of some Polish space $\tilde X$. Let $\{V_n\}_{n\in\omega}$ be a countable base of the topology of $Y$. Since $f$ is Borel, for every $n\in\omega$ the preimage $f^{-1}(V_n)$ is Borel in $X$ and hence $f^{-1}(V_n)=X\cap B_n$ for some Borel set $B_n\subset\tilde X$.

By a known result of Descriptive Set Theory (see Exercise 13.5 in ``Classical Descriptive Set Theory'' of A.Kechris), there exists a continuous bijective map $g:\tilde Z\to \tilde X$ from a Polish space $\tilde Z$ such that $g^{-1}(B_n)$ is clopen for any $n\in\omega$. Consider the set $Z=g^{-1}(X)$ and observe that the map $h=f\circ g{\restriction}Z$ is continuous, and hence can be entended to a continuous map $\tilde h:G\to Y$ defined on some $G_\delta$-set $G\subset\tilde Z$, containing $Z$.

Since $|Z|=|g^{-1}(X)|=|X|<\mathfrak b$, the set $Z=g^{-1}(X)$ is contained in some $\sigma$-compact subset $A=\bigcup_{n\in\omega}A_n$ of $G$. For every $n\in\omega$ the set $K_n:=g(A_n)\subset \tilde X$ is compact and $g{\restriction}A_n:A_n\to K_n$ is a homeomorphism. Then the restriction $f{\restriction}K_n\cap X=f\circ g\circ (g{\restriction}A_n\cap Z)^{-1}$ is continuous, and $f$ is $\bar \sigma$-continuous. $\quad\square$

Theorem U implies:

Corollary U. Any function $f:X\to Y$ from a submetrizable $Q$-space $X$ of cardinality $|X|<\mathfrak b$ to a metrizable separable space $Y$ is $\bar\sigma$-continuous.

In its turn, Corollary U implies the following answer to the second part of Problem 3.

Corollary. $\bar\sigma=\mathfrak q_0=\min\{\sigma,\mathfrak b,\mathfrak q\}$.

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