# Path-continuity and the Axiom of Choice

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.

• BTW, I got 𝑓∘𝛾 by copying and pasting. What is the LaTeX for the centered circle? – Stephen Herschkorn Mar 29 '19 at 4:14
• $\circ$ is \circ – Alex Kruckman Mar 29 '19 at 4:25
• 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 $a$. – Eric Wofsey Mar 29 '19 at 5:11
• 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. – Eric Wofsey Mar 29 '19 at 5:31
• 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 $\infty$). – Eric Wofsey Mar 29 '19 at 5:46

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 $$(Y_m)_{m. Then $$(*)$$ gives a choice function $$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.

• 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. – Stephen Herschkorn Mar 29 '19 at 5:22
• 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.) – Eric Wofsey Mar 29 '19 at 5:26
• Oh, you're right about the relevant set. I think I'll fix it in the original post. – Stephen Herschkorn Mar 29 '19 at 6:11
• @EricWofsey Don't you think that any analytic set existence require AC ? – Soleil Mar 29 '19 at 9:25
• ... aha, I see it now. You can use the "same" curve each time, scaled to the different balls. – David Hartley Mar 29 '19 at 14:19