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It is well-known that ZF cannot prove the following:

"for a function $f$ from reals to reals and any real $x$, $f$ is continuous at $x$ if and only if $f$ is sequentially continuous at $x$."

See e.g. Herrlich's excellent monograph "Axiom of Choice".

What kind of restrictions can one impose on $f$ while keeping the unprovability in ZF?

I am talking about restrictions in the sense of function classes, i.e. semi-continuity or Baire 2.

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    $\begingroup$ Let me verify that I understand you correctly. You are asking about a class $C$ of functions, such that the following is not provable in ZF: $\forall f \in C.\, \forall x \in \mathbb{R} .\, (\mathrm{continuous}(f,x) \Leftrightarrow \mathrm{sequentiallyContinuous}(f,x))$. We know that ZF does not prove the statement when $C$ is the class of all functions, but you'd like a smaller class. Correct? $\endgroup$
    – Andrej Bauer
    Commented Oct 1 at 20:40
  • $\begingroup$ @AndrejBauer That is exactly what I mean. $\endgroup$
    – Sam Sanders
    Commented Oct 2 at 9:02

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If the graph of $f$ has a Borel code $c,$ then ZF proves equivalence of these two continuity concepts. Discontinuity of $f$ at $x$ is downward absolute to $L[c, x],$ in which we can find a sequence $x_i \rightarrow x$ for which it is not the case that $f(x_i) \rightarrow f(x).$

Note that semi-continuous $\rightarrow$ Baire class 1 $\rightarrow$ Baire class 2, and every Baire class 2 function has a Borel code, see: The difference between Baire 2 and 'effectively Baire 2'. So no counterexamples can occur in the cases you list.

Consistently, the equivalence fails for Baire class 3. If $\mathbb{R}$ is a countable union of countable sets $X_n,$ then every set of reals has a BC3 indicator function. Now let $x=0,$ and let $f$ be the indicator function for $\bigcup_{n<\omega} \{y \in [2^{-n-1}, 2^{-n}]: \forall i<n (y \text{ codes an enumeration of } X_i)\}.$ This is discontinuous at $x,$ but a failure of sequential continuity at $x$ would provide a way to enumerate $\mathbb{R},$ so that can't happen.

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  • $\begingroup$ Thanks for the nice answer and counterexample. It seems that the same does not hold for the sequential definition of semi-continuity, i.e. for sequentially usco functions, the local equivalence between continuity and sequential continuity, is not provable in ZF. $\endgroup$
    – Sam Sanders
    Commented Oct 5 at 19:42
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    $\begingroup$ Yes, that fails for the indicator function of an infinite Dedekind finite set $S:$ globally sequentially usco, sequentially continuous off $S,$ discontinuous at the condensation points of $S$ (which is a perfect set so is not contained in $S$). If you demanded $f$ be in some Baire class AND usco, it would be significantly harder to find a counterexample. $\endgroup$
    – Elliot Glazer
    Commented Oct 6 at 0:28
  • $\begingroup$ Indeed, while usco functions are Baire 1, the latter is "hard to prove" at least in higher-order arithmetic. $\endgroup$
    – Sam Sanders
    Commented Oct 14 at 20:51

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