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.

differentletter, 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 '19 at 11:30