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Stream: deprecated: mathematics

Topic: The closure functor

Jean-Baptiste Vienney (Jun 12 2023 at 17:59):

Let $Bounded$ be the category whose objects are bounded subsets of euclidean spaces $\mathbb{R}^{n}$ for any $n \ge 1$ and whose morphisms are Lipschitz continuous functions. For every $A \in Bounded$, $\overline{A}$ is a closed and bounded metric space with the induced distance, thus compact and thus complete.

Let $f:A \rightarrow B$, and $x \in \overline{A}$, let $(x_{n})_{n \ge 0}$ be a sequence of points of $A$ which converges to $x$ in $\overline{A}$.

For every $n,p \ge 0$, we have that $d(f(x_{n}),f(x_{p})) \le K.d(x_{n},x_{p})$ where $K \ge 0$ is the Lipschitz constant of $f$. $(x_{n})_{n \ge 0}$ is a convergent sequence in $\overline{A}$ and thus a Cauchy sequence in $\overline{A}$ and thus a Cauchy sequence in $A$. It follows from this inequality that $(f(x_{n}))_{n \ge 0}$ is also a Cauchy sequence in $B$ and thus a Cauchy sequence in $\overline{B}$ and thus converges in $\overline{B}$ because it is complete. Moreover this limit doesn't depend on the choice of sequence $(x_{n})_{n \ge 0}$. We can thus write it $\overline{f}(x) \in \overline{B}$ as it only depends on $x$. We then obtain a function $\overline{f}:\overline{A} \rightarrow \overline{B}$ that we can show is continuous.

I believe that we obtain a closure functor $\bar{.}:Bounded \rightarrow Bounded$. It looks like a very useful functor -- people who use applied functional analysis use the type of reasoning described above everyday but it would probably be easier for them if they could directly use such functors instead of redoing this kind of proofs all the time. I didn't find references about this. There is a book Topology: A Categorical Approach but it is not that much categorical.

Jean-Baptiste Vienney (Jun 12 2023 at 18:07):

If anybody knows a reference, please tell me. Otherwise I believe it could make a nice half-page paper.

Jean-Baptiste Vienney (Jun 12 2023 at 18:15):

Some years ago, my first research project was with a prof working in functional analysis and on such things as fluid mechanics and the Navier-Stokes equation. I was doing such reasoning like this all the time with her. I could ask her about this functor! I think she knows nothing about category theory, but she could maybe be excited to know learn that this thing is a functor!

Jean-Baptiste Vienney (Jun 12 2023 at 18:25):

But maybe it is something basic in modal logic where they have closure operators like this?

Mike Shulman (Jun 12 2023 at 18:28):

Is this "just" the completion functor on metric spaces, restricted to bounded subsets of $\mathbb{R}^n$?

Jean-Baptiste Vienney (Jun 12 2023 at 18:31):

Probably yes, I just take the closure of a bounded subset of $\mathbb{R}^{n}$ and it gives a complete metric space .

Jean-Baptiste Vienney (Jun 12 2023 at 18:32):

So that's nothing more than that I guess.

Jean-Baptiste Vienney (Jun 12 2023 at 18:33):

Hmm I don't know because I need Lipschitz functions.

Jean-Baptiste Vienney (Jun 12 2023 at 18:33):

I don't see how you could extend your function without it being Lipschitz?

Mike Shulman (Jun 12 2023 at 18:41):

Well, you could just take the category of metric spaces and Lipschitz maps, if you want your category to be a full subcategory of it. But it suffices to use uniformly continuous maps (in which case you can also generalize the objects to [[uniform spaces]]).

Jean-Baptiste Vienney (Jun 12 2023 at 18:46):

Ok perfect, I've found a paper which talk about a completion functor. So I'm happy to know about this functor which is already well-known!