Choiceless vs choicy proof

Give an illustration of how a choiceless proof can be more perspicuous than a choicy one.

Nice question — this gets to the heart of why mathematicians sometimes prefer avoiding the Axiom of Choice: a choiceless (i.e. choice-free or constructive) proof can actually explain how and why something happens, not just assert that it does.

Below I give a concrete illustration: a short, explicit choiceless proof of compactness for countable products of compact metric spaces (a very classical diagonal/subsequence argument) and then contrast it with the standard choicy proof of the full Tychonoff theorem (which uses ultrafilters/Zorn/choice). The contrast shows why the choiceless proof feels more perspicuous.

Example — countable product of compact metric spaces (choiceless, very perspicuous)

Theorem. If $(X_n,d_n)$ are compact metric spaces for $n\in\mathbb{N}$, then the product $X=\prod_{n\in\mathbb{N}} X_n$ (with the product metric, say $d(x,y)=\sum_{n=1}^\infty 2^{-n}\min(1,d_n(x_n,y_n))$) is compact.

Proof (diagonal subsequence — fully explicit). Let $(x^{(m)})_{m\in\mathbb{N}}$ be any sequence in $X$; write $x^{(m)}=(x^{(m)}_1,x^{(m)}_2,\dots)$. We produce a convergent subsequence of $(x^{(m)})$ by successive extraction:

1. Because $X_1$ is compact, the sequence $(x^{(m)}1){m}$ has a convergent subsequence. Choose an increasing sequence of indices $m^{(1)}_1<m^{(1)}_2<\cdots$ such that $x^{(m^{(1)}_k)}_1$ converges in $X_1$.

2. Look at the subsequence of whole points $(x^{(m^{(1)}k)}){k}$. Since $X_2$ is compact, the sequence of second coordinates $(x^{(m^{(1)}_k)}_2)_k$ has a convergent subsequence. Choose a further subsequence of indices $m^{(2)}_1<m^{(2)}_2<\cdots$ (each $m^{(2)}j$ is some $m^{(1)}{k}$) such that $x^{(m^{(2)}_j)}_2$ converges.

3. Continue inductively: at stage $n$ we have an increasing sequence of indices $m^{(n)}_1<m^{(n)}_2<\cdots$ so that for each $i\le n$ the $i$-th coordinates $x^{(m^{(n)}_k)}_i$ converge in $X_i$.

4. Define the diagonal subsequence by $m_k := m^{(k)}_k$. This is an increasing sequence of indices. Fix any coordinate $i$. For all $k\ge i$, the index $m_k=m^{(k)}k$ belongs to the sequence $m^{(i)}\bullet$ (because the sequences were nested), so the $i$-th coordinate $x^{(m_k)}_i$ tends to the same limit chosen at stage $i$. Hence for each fixed $i$ we have $x^{(m_k)}_i\to \ell_i$ as $k\to\infty$.

5. Coordinatewise convergence plus the definition of the product metric implies $x^{(m_k)}\to \ell=(\ell_1,\ell_2,\dots)$ in $X$. Thus $(x^{(m)})$ had a convergent subsequence, so $X$ is sequentially compact; in metric spaces (and first countable spaces) sequential compactness equals compactness, so $X$ is compact.

This proof is explicit and algorithmic: given any sequence you literally describe how to extract a convergent subsequence. It reveals the mechanism (diagonal extraction) and even can be implemented as a procedure.

The choicy route (less perspicuous)

The full general Tychonoff theorem — an arbitrary product of compact spaces is compact — is typically proved by one of these choice-heavy methods:

• via ultrafilters: every ultrafilter on the product projects to ultrafilters on the factors; compactness of factors gives limits there; one reassembles them and uses the existence of ultrafilters etc.; or
• via Zorn’s lemma / maximal families with the finite intersection property.

Both approaches are extremely powerful but non-constructive: they assert existence of limit points without giving a recipe to construct them, and they rely on the Axiom of Choice (or an equivalent). For many readers the ultrafilter/Zorn proof is short at the level of set theory but opaque — it hides the concrete mechanism of how convergence in the product actually looks coordinatewise and gives no algorithm for producing convergent subsequences even in the countable case.

Why the choiceless proof is more perspicuous

• Constructiveness / algorithmic content. The diagonal proof tells you how to build the limit; the ultrafilter proof only tells you that a limit exists. If you want an explicit subsequence, the choiceless diagonal method gives it.

• Mechanism revealed. The diagonalization shows the product compactness “comes from” compactness of individual coordinates, one coordinate at a time. The ultrafilter/Zorn argument packages that into an abstract existence principle and obscures the step-by-step idea.

• Better for intuition and applications. If you need to estimate rates, extract quantitative information, or implement the construction (e.g. in analysis or numerical methods), the constructive diagonal argument is usable; the choice proof is not.

• No heavy set theory needed. The choiceless proof can be checked by elementary real-analysis style reasoning, which is often clearer to readers not steeped in axiomatic set theory.

A short moral (and a caveat)

Choiceless proofs are often more perspicuous because they are constructive and expose the underlying mechanism. But there are important theorems (like full Tychonoff, or “every vector space has a basis”) that generally do require choice for full generality — and in those cases the choicy proof is sometimes the only available route. Often the right attitude is: try to prove as much as you can without choice (you may get a clearer, more informative proof), and only call in choice when the problem truly needs it.

If you want, I can: • Give another pair of proofs (e.g. existence of bases in finite vs. infinite dimension), or • Formalize the diagonal proof in full detail for a particular product metric, or • Show a worked example extracting a subsequence from a concrete sequence in $[0,1]^\mathbb{N}$. Which would you like?