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Stream: learning: questions

Topic: math.stackexchange questions to resolve

Ryan Schwiebert (Jan 08 2026 at 14:23):

I plan to drop links to select category theory math.se questions that haven’t been resolved yet, in case an attendee here is not on that site.

Ryan Schwiebert (Jan 08 2026 at 14:23):

The comment here sounds like a challenge to me:
https://math.stackexchange.com/questions/5118756/is-there-a-categorical-view-point-for-lim-sup-and-lim-inf

James Deikun (Jan 08 2026 at 15:25):

It's not clear to me exactly what is meant by "lim sup and lim inf of functions" in the question, if that is clarified maybe there could be an answer.

Ryan Schwiebert (Jan 08 2026 at 15:31):

My impression was the function computed point-wise as described here: https://en.wikipedia.org/wiki/Limit_inferior_and_limit_superior#Real-valued_functions

Ruby Khondaker (she/her) (Jan 08 2026 at 16:10):

The universal property I'm familiar with for supremum is that $x \geq \sup S \iff \forall s \in S, x \geq s$. As far as I'm aware, there's nothing similar of this type for limsup - for example, $\limsup_{n \to \infty} \frac 1n = 0$, but $0$ is strictly less than every element of the sequence!

However, there is a nice alternative way to look at this. $x \leq \sup S \iff ([y \geq \sup S] \implies [y \geq x])$ by Yoneda. And that's $\iff ([\forall s \in S, y \geq s] \implies [y \geq x])$. In other words, $x \leq \sup S$ if and only if every upper bound of $S$ is $\geq x$. If we view the real numbers as a poset under $\leq$, then you can view this under the guise of isbell duality - we've turned $\sup S$ from a covariant set-valued functor to a contravariant set-valued functor.

In this sense, there is an analogous universal property for limsup. $x \leq \limsup S \iff ([\forall^\infty s \in S, y \geq s] \implies [y \geq x])$, where $\forall^\infty$ means "for all but finitely many". In other words, $x \leq \limsup S$ if and only if every essential upper bound of $S$ (meaning an upper bound for "almost all" elements of S) is $\geq x$.

Unlike the supremum case, $\forall^\infty s \in S, y \geq s$ is not necessarily representable. It is true that $[\forall^\infty s \in S, y \geq s] \implies y \geq \limsup S$, but the reverse is no longer true by considering the $\frac 1n$ example.

I'm not aware of a nice categorical way to talk about something holding "almost everywhere" - this seems to be a fundamentally analytic concept. Nevertheless, this feels like the closest thing to a "universal property" for limsup. And dually for liminf.

I'd be curious to hear what other people think about this!

James Deikun (Jan 08 2026 at 16:15):

The characterization of limsup using $\forall^\infty s \in S, y \geq s$ only works for sequences, not for functions from reals to reals.

Ruby Khondaker (she/her) (Jan 08 2026 at 16:16):

Yes I haven't thought about how this works for functions just yet

Ryan Schwiebert (Jan 08 2026 at 16:18):

There’s a chance the user might have been thinking of limsup/liminf of a sequence of functions I suppose. Asking for that clarification

James Deikun (Jan 08 2026 at 16:33):

The essential concepts being used in the definition referenced by @Ryan Schwiebert are the set of neighborhoods of a point in a topological space, and a restriction of a partial function. Then lim sup is defined as being a lower bound for all upper bounds of restrictions of $f$ to neighborhoods of the point in question. This can be seen as a colimit of limits, but I wonder if it's useful to do so.

Ruby Khondaker (she/her) (Jan 13 2026 at 10:07):

Unsure if this is still relevant, but after a bit more pondering I figured out some alternative universal properties for limsup and liminf.

$\limsup S \leq M \iff \forall \epsilon > 0, \forall^\infty s \in S, s \leq M + \epsilon$, and dually $m \leq \liminf S \iff \forall \epsilon > 0, \forall^\infty s \in S, m - \epsilon \leq s$. So they capture "eventual" behaviour of your sequence up to an $\epsilon$ of room.