It is well-known that for a product space $\Pi_{\alpha \in A} X_{\alpha}$, a sequence of points $\{\mathbf{x}_n\}$ converges if and only if for each $\alpha \in A$, $\{\pi_{\alpha}(\mathbf{x}_n)\}$ converges in $X_{\alpha}$. The traditional proof relies heavily on arguments of neighborhoods, and it is in some sense long and laborious.

My question is that, since the product space is a product in categorical sense, would a depiction of convergence in categorical sense neatly prove the assertion?


1 Answer 1


Yes. A sequence $\{x_k\}$ that converges to $x_\infty$ is nothing but a continuous function from the topological space $\mathbb N_\infty$ (the natural numbers plus a point $\infty$ with the subspace topology from the extended real numbers), where $x_k$ is the function value at $k$.

Therefore, a sequence $p:\mathbb N \to X$ converges iff there is a continuous map $\tilde p :\mathbb N_\infty \to X$ such that $p = \tilde p \circ \iota$ where $\iota$ is the obvious inclusion map. We have now completely rewritten the problem with category theory, and from now on no more topology is needed. Can you now prove it with pure abstract nonsense? (I can expand the answer if you have difficulties.)

  • $\begingroup$ Then since $\Pi_{\alpha} Hom (\mathbb{N}_{\infty}, X_{\alpha} ) \cong Hom (\mathbb{N}_{\infty}, \Pi_{\alpha} X_{\alpha} )$ it easily follows the result! $\endgroup$ Dec 1, 2023 at 4:12
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    $\begingroup$ The cool thing is that you can do the same with convergent nets by considering $P \cup \{\infty\}$ for directed sets $P$, thus recovering the whole topology of the space. This observation is the basis of my paper on topological space objects that might be of interest for you. arxiv.org/abs/2106.11115 $\endgroup$ Dec 1, 2023 at 7:58

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