Two days ago, I wrote about the construction of a global section $s$ of a stable projection $\pi: W \twoheadrightarrow V$ from local sections $s_v: V \cap U_v \to W$ which cover the entire $V$. A crucial formula was

$s(\tilde v) := \sum_{v \in V} \rho_v(\tilde v) s_v(\tilde v),$

describing the global section as a convex combination of locally finitely many local sections using a partition of unity $\rho_v$ subordinate to the open cover $(U_v)_{v \in V}$.

The formula in that article now looks somewhat different, because upon reading the post, Andreas (who, if I remember correctly, was the one to first write some version of this formula on the blackboard when we were discussing stable equivalence) remarked that the original formulation reproduced above was misleading because the sum runs over $V$ and not the locally finite refinement of the cover that I was briefly hinting at in the paragraph before. I did this deliberately to avoid having to interrupt the main line of the proof.

But I was caught, so here is that interruption.

Suppose local sections $s_v: V \cap U_v \to W$. Then the sets $(U_v \cap V)_{v \in V}$ are an open cover of the primary basic semialgebraic set $V$. Since $V$ is paracompact, there exists a locally finite refinement $(U'_i)_{i \in I}$ of this cover on $V$. Local finiteness is sufficient to give $s(\tilde v)$ a well-defined value and make it continuous, but with the new index set $I$, over which I know a priori nothing, I am unable to write down the convex combination of local sections. Andreas remedies this in his email by invoking the axiom of choice to select a map $f: I \to V$ such that $U'_i \subseteq U_{f(i)}$ and all is well because we can now sum correctly:

$s(\tilde v) := \sum_{i \in I} \rho_i(\tilde v) s_{f(i)}(\tilde v).$

All is well, *except* he includes a small comment

(in general you probably need the axiom of choice)

Now, I am not against choice, but I do try to avoid it whenever I can. The feeling I have is that you need choice if you don’t know your set well enough to make a choice function. This may be true for an anonymous paracompact space. I would be more reluctant to accept that choice is strictly required to do this in a second-countable metric space (if you know, please tell me). But I really like primary basic semialgebraic sets in $\mathbb{R}^n$ and I don’t want it to look like I don’t know them well enough.

To understand my relation to the axiom of choice, let me take you back 9 years in time. After I finished school, I was trying to decide what to study: Mathematics or Computer Science. Somehow I had heard that “sets” would be important in university mathematics and so I bought the first set theory book with the friendliest title I could find, Oliver Deiser: *Einführung in die Mengenlehre*. **This is a truly wonderful book.** It taught me, a self-studying high school graduate without anyone to talk to, how to think mathematically and how to prove things (because I did all the exercises!). It was fun, engaging and contained interesting bits of historical information. Needless to say I chose mathematics and was well prepared for the first semester. My highest recommendation to any German-reading future math student.

The book made a point of emphasizing the usage and necessity of its invocations of the axiom of choice and this has been imprinted on me simply because the book had such an influence on my early mathematical development.

Luckily, there is a mathematical discipline for making choice functions. It has MSC2020 numbers in the range 90-XX and is called Optimization. Whenever I try to avoid the axiom of choice, I try to formulate an optimization problem whose solutions replace the anonymous choice function.

Suppose we have an open cover $(U_v)_{v \in V}$ of $V$ consisting of neighborhoods $U_v \ni v$. The goal is to thin this cover out and find a subset $V' \subseteq V$ such that $(U_v)_{v \in V'}$ is a locally finite cover of $V$. “Thinning out”, i.e., making a subcover, is not quite what we will achieve below. Instead, we make each $U_v$ smaller first and then select a subcover. In any case, we obtain a locally finite refinement which retains the property that its open sets $U'_v$ are indexed by points $v \in U'_v$.

Since $V$, as a primary basic semialgebraic set, is locally closed in $\mathbb{R}^n$, it is (semialgebraically) homeomorphic via $f$ to a closed set in $\mathbb{R}^{n+1}$. This is probably a standard construction in metric topology (and also in the Zariski topology!) which encodes $V$ by the points in its closure which have a non-zero distance to the boundary. The latter condition can be formulated in a closed way by introducing an additional variable. I found it in the special case of semialgebraic geometry in Bochnak-Coste-Roy’s book *Real algebraic geometry*, Proposition 2.2.9. The transition to $f(V)$ changes nothing about our objective: open covers can be transferred through the homeomorphism and local finiteness is preserved because it is just a set-theoretic property.

At this point we would like to very slightly refine the cover $(U_v)_{v \in V}$ of $V$ to ensure that the image cover satisfies $f(U_v) \subseteq B_1(f(v))$. This can be done by simply replacing $U_v$ with $f^{-1}\left[f(U_v) \cap B_1(f(v))\right]$. Since $(U_v)_{v \in V}$ is such a dense cover — it has a neighborhood of every point —, this operation gives another open cover.

To get a subcover of this modified $(U_v)_{v \in V}$, we work through countably many compact slices $C^{(r)} := f(V) \cap \bar{B}_r(0)$ where $\bar{B}_r(0)$ is the closed ball of radius $r \in \mathbb{N}$. Whenever $C^{(r)}$ is exhausted by the below procedure, we increment $r$ and continue for as long as $f(V)$ is not completely exhausted. Set $C^{(r)}_0 := C^{(r)}$ and $k := 0$ and iteratively do the following:

Pick that $v \in C^{(r)}_k$ with the lexicographically smallest coordinates and add it to $V'$. (This is our choice function and it is well-defined because $C^{(r)}_k$ is compact.)

Set $C^{(r)}_{k+1} := C^{(r)}_k \setminus f(U_v)$ which is still a compact set and go to step 1 with $k := k+1$.

This procedure exhausts $C^{(r)}$ after a finite amount of steps because the images $f(U_v)$ which cover $f(V)$ in $\mathbb{R}^{n+1}$ have a certain positive Lebesgue measure (which is also finite under the intersection with $\bar{B}_r(0)$) which is depleted in every step.

The result of this procedure is a countable subset $V' \subseteq V$ such that $(f(U_v))_{v \in V'}$ is an open cover of $f(V)$ and hence $(U_v)_{v \in V'}$ covers $V$. Moreover, this cover is locally finite: for every point $v$ chosen in the above procedure, its neighborhood is removed before choosing the next point (which prevents accumulation points); and since all $f(U_v)$ are contained in a unit ball around $f(v)$, only the finitely many other chosen neighborhoods from at most three consecutive slices $C^{(r)}$ may intersect any given neighborhood.