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This question already has an answer here:

Does there exist a model of $ZF¬C$ in which there is a function $f:\mathbb R \to \mathbb R$ such that $f$ is sequentially continuous at some $a \in \mathbb R$ but not $\epsilon-\delta$ continuous , i.e., for any sequence $\{x_n\}$ converging to $a$ , $\lim f(x_n)=f(a)$ but still $f$ is not continuous at $a \in \mathbb R$ ?

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marked as duplicate by Asaf Karagila axiom-of-choice Jun 18 '15 at 4:43

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This is certainly possible for a specific point: Suppose there is a infinite, Dedekind-finite set of reals $A$. Then let $f$ be the characteristic function of $A$. Since $A$ cannot be closed (perfect set property), let $a$ be in the closure of $A$ but not in $A$. Then $f$ is sequentially continuous at $a$ (since no $\omega$-sequence of elements of $A$ exists), but not continuous at $A$. In fact, $f$ is sequentially continuous at exactly the points not in $A$, and continuous at exactly the points not in the closure of $A$.

(Of course, it's straightforward to show in ZF alone that any $0-1$ valued function which is sequentially continuous is continuous.)


However, in ZF alone, if $f$ is sequentially continuous everywhere then it must actually be continuous.

Suppose $f$ is sequentially continuous everywhere but discontinuous at $a$. Since $\mathbb{Q}$ is dense and well-orderable, for each $\epsilon>0$ there must be a $\delta>0$ such that $$\forall q\in\mathbb{Q}(\vert q-a\vert<\delta\implies \vert f(q)-f(a)\vert<\epsilon).$$ Fix a function $m$ mapping each $\epsilon$ to an appropriate $\delta$ (say, by taking the sup of the $\delta$s that work and dividing by $2$). Say that a real $r$ is bad if $\vert r-a\vert<m(\vert {f(r)-f(a)\over 2}\vert)$. Such a real $r$ must exist since $f$ is continuous at $a$. But now any sequence of rationals approaching $r$ witnesses the failure of sequential continuity.


What remains open: it is not clear to me that "sequentially continuous at a point implies continuous at that point" is actually equivalent to "there is no Dedekind-finite infinite set of reals," which is strictly weaker than countable choice for reals (which is the obvious upper bound).

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    $\begingroup$ It’s equivalent to $\operatorname{CC}(\Bbb R)$; see my answer to the OP’s earlier question. $\endgroup$ – Brian M. Scott Jun 18 '15 at 4:23