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We tell the following to our Calc III students (usually for $mathbf{R}^2$, and never so formally):

Let $A$ be an open subset of $mathbf{R}^n$, $ain A$, $f$ a real-valued function on $A$ and $Gamma = {gamma in A^{[0,1]} : gamma(0) = a$ and $gamma$ is continuous at $0}$. If 𝑓∘𝛾 is continuous at $0$ for all $gamma in Gamma$, then $f$ is continuous at $a$.

We may generalize this: $A$ can be a first countable locally path-connected space, and the codomain may be any topological space. See Continuity on paths implies continuity on space? .

Here is my question:

Does one need the Axiom of Choice (or at least countable choice) to prove this result?

For example, the essentials of my proof of the (restricted) result is: If $f$ is not continuous at $a$, then there is a neighborhood $V$ of $f(a)$ such that $f^{-1}(V)$ is not a neighborhood of $a$. For each positive integer $n$, choose a point in $ B(a,1/n)backslash f^{-1}(V)$ and connect the points with a piecewise linear path. Thus, we are choosing countably many points.

The proof at the link above also uses countable choice.

For open subsets $Asubseteqmathbb{R}^n$, no choice is needed. Indeed, note that if $Bsubsetmathbb{R}^n$ is any closed ball, there is a continuous surjection from an interval to $B$. Concatenating infinitely many such space-filling curves onto smaller and smaller balls around a point $a$, we get a single continuous path $gamma:[0,1]to A$ such that a map $f$ on $A$ is continuous at $a$ iff $fcircgamma$ is continuous at $0$.

Let us now discuss more general spaces; the upshot is that the axiom of choice is required to prove these results in general. I will say a space $A$ is pseudopath-generated if it satisfies your condition: that is, a map $f:Ato Y$ is continuous at a point $ain A$ iff for any map $gamma:[0,1]to A$ such that $gamma(0)=a$ and $gamma$ is continuous at $0$, $fcircgamma$ is continuous at $0$. I will say a space $A$ is path-generated if the same condition holds but with $gamma$ required to be continuous on all of $[0,1]$.

Note that the result you linked regarding first countable locally path-connected spaces actually says that they are path-generated, not just pseudopath-generated. Indeed, pseudopath-generation doesn't really have anything to do with paths, since continuity of $gamma$ at a single point is a very weak condition. A similar argument shows that in fact every first countable space is pseudopath-generated.

The statement that every first countable locally path-connected space is path-generated is equivalent to countable choice. Let me first observe that countable choice is equivalent to the following (seemingly weaker) statement.

$(*)$ Given a countable family of $(X_n)_{ninmathbb{N}}$ of nonempty sets, there exists an infinite subset $Ssubseteqmathbb{N}$ and a choice function for $(X_n)_{nin S}$.

To prove countable choice from $(*)$, let $(Y_n)_{ninmathbb{N}}$ be a countable family of nonempty sets and let $X_n$ be the set of choice functions for the initial segment $(Y_m)_{m<n}$. Then $(*)$ gives a choice function $f$ on an subsequence of $(X_n)$, which then gives a choice function $g$ for all of $(Y_n)$ (let $g(m)=f(n)(m)$ where $n$ is minimal in the domain of $f$ such that $f(n)(m)$ is defined).

Now here is a sketch of a proof that path-generation of first countable locally path-connected spaces implies $(*)$. Let $(X_n)$ be a countable family of nonempty sets. Let $*$ be some point that is not an element of any $X_n$ and let $Y_n=X_ncup{*}$. Let $G$ be the graph with vertex set $mathbb{N}$ and an edge from $n$ to $n+1$ for each element of $Y_n$. Consider this graph as a topological space, not with the weak topology but with the natural metric that makes each edge a path of length 1. Let $A=Gcup{infty}$, where a neighborhood of $infty$ must contain all sufficiently large vertices and edges between them in $G$.

Then $A$ is first countable and locally path-connected (for local path-connectedness at $infty$, we use the fact that we can get a path from any vertex of $G$ to $infty$ by infinitely concatenating the paths from $n$ to $n+1$ indexed by $*$). Now consider the map $f:Ato[0,1]$ defined as follows: $f(x)=0$ if $x$ is a vertex of $G$ or $infty$, or is on one of the paths labelled by $*$. On each path labelled by an element of $X_n$, $f$ starts at $0$, goes up to $1$, and then goes back down to $0$.

This map $f$ is not continuous at $infty$, since every neighborhood of $infty$ contains paths labelled by elements of $X_n$. But to witness this discontinuity with a path in $A$, we would need a path $gamma:[0,1]to A$ which passes through paths corresponding to elements of $X_n$ for arbitrarily large $n$. Such a path would give a choice function on a subsequence of $(X_n)$ (for instance, for each $n$ such that $gamma$ goes into the path corresponding to an element of $X_n$, choose the least rational point in $[0,1]$ with respect to some well-ordering of the rationals which maps into the path corresponding to an element of $X_n$, and use that to choose an element of $X_n$).

Let me now return to pseudopath-generated spaces. Everything in the argument above still works if $A$ is known only to be pseudopath-generated, except for the very last step where we use the path $gamma$ to get a choice function on a subsequence of $(X_n)$. If $gamma$ is not assumed to be continuous on $[0,1]$, then we do not have a canonical way to choose one particular element of $X_n$ which it hits. However, if we assume that $mathbb{R}$ can be well-ordered, then the argument still works and proves countable choice. In particular, since ZF+"there exists a well-ordering of $mathbb{R}$" cannot prove countable choice, this implies that ZF cannot prove every first countable locally path-connected space is pseudopath-generated.

By similar arguments we can prove the following are equivalent over ZF:

1. All first-countable spaces are pseudopath-generated.


2. If $(X_n)$ is any family of nonempty sets then there exists a function $f$ on $mathbb{N}timesmathbb{R}$ such that for each $n$, there exists $rinmathbb{R}$ such that $f(n,r)in X_n$.

Statement 2 follows from countable choice and cannot be proved in ZF, but I suspect it is strictly weaker than countable choice.