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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 $\mathbf{R}^n$, $a\in A$, $f$ 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.

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  • $\begingroup$ BTW, I got 𝑓∘𝛾 by copying and pasting. What is the LaTeX for the centered circle? $\endgroup$ – Stephen Herschkorn Mar 29 '19 at 4:14
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    $\begingroup$ $\circ$ is \circ $\endgroup$ – Alex Kruckman Mar 29 '19 at 4:25
  • $\begingroup$ I have realized that I wrote my answer assuming that elements of $\Gamma$ are required to be continuous on all of $[0,1]$, not just at $0$. If you only require them to be continuous at $0$ then the result holds in every first-countable space (no local path-connectedness needed), since you can just take $\gamma$ to be a piecewise function taking values in an arbitrary sequence approaching $a$. $\endgroup$ – Eric Wofsey Mar 29 '19 at 5:11
  • $\begingroup$ In any case, with the definition you wrote (which I think is considerably less interesting than the version that requires $\gamma$ to be continuous everywhere), some choice is still needed. But I think it is strictly less than countable choice; in particular, it suffices to know that given a sequence $(X_n)$ of nonempty sets, there exists a subsequence and a simultaneous choice of a function from some nonempty subset of $\mathbb{R}$ to each term in the subsequence. This seems strictly weaker than countable choice. $\endgroup$ – Eric Wofsey Mar 29 '19 at 5:31
  • $\begingroup$ Note also that the question you linked asks a global question, about continuity on the entire space $A$ instead of continuity at a single point. The example space in my answer still works for that global statement (since $f$ is continuous at every point other than $\infty$). $\endgroup$ – Eric Wofsey Mar 29 '19 at 5:46
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For open subsets $A\subseteq\mathbb{R}^n$, no choice is needed. Indeed, note that if $B\subset\mathbb{R}^n$ 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 $f\circ\gamma$ 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 $f:A\to Y$ is continuous at a point $a\in A$ iff for any map $\gamma:[0,1]\to A$ such that $\gamma(0)=a$ and $\gamma$ is continuous at $0$, $f\circ\gamma$ 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 $(X_n)_{n\in\mathbb{N}}$ of nonempty sets, there exists an infinite subset $S\subseteq\mathbb{N}$ and a choice function for $(X_n)_{n\in S}$.

To prove countable choice from $(*)$, let $(Y_n)_{n\in\mathbb{N}}$ 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 $Y_n=X_n\cup\{*\}$. 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 $A=G\cup\{\infty\}$, 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 $f:A\to[0,1]$ 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 $\mathbb{N}\times\mathbb{R}$ such that for each $n$, there exists $r\in\mathbb{R}$ 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.

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  • $\begingroup$ For the $\mathbf{R}^n$ case, how do we know $f^{-1}(V) \backslash B(a,1/n)$ has rational points? I pose this while fearing I am missing something obvious. $\endgroup$ – Stephen Herschkorn Mar 29 '19 at 5:22
  • $\begingroup$ Oh, oops, I was just wrong. (By the way, the relevant set is $B(a,1/n)\setminus f^{-1}(V)$, not the other way around.) $\endgroup$ – Eric Wofsey Mar 29 '19 at 5:26
  • $\begingroup$ Oh, you're right about the relevant set. I think I'll fix it in the original post. $\endgroup$ – Stephen Herschkorn Mar 29 '19 at 6:11
  • $\begingroup$ @EricWofsey Don't you think that any analytic set existence require AC ? $\endgroup$ – Soleil Mar 29 '19 at 9:25
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    $\begingroup$ ... aha, I see it now. You can use the "same" curve each time, scaled to the different balls. $\endgroup$ – David Hartley Mar 29 '19 at 14:19

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