Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: ε-δ is "semi-adjunction"?

Joshua Meyers (Feb 21 2021 at 22:45):

Let $f:\mathbb{R}\to\mathbb{R}$ be a function. Consider the statement that $f$ is continuous at $x_0\in\mathbb{R}$. We can define a "modulus of continuity" $\omega:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ by $\omega(\delta)\coloneqq \text{sup}\{|f(x)-f(x_0)|:|x-x_0|< \delta\}$. Then $f$ is continuous at $x_0$ iff $\omega$ is continuous at $0$. Also note that $\omega$ is a weakly monotonic function (i.e. it preserves the relation $\leq$). So we have reduced the notion of continuity of a real function at a point to that of continuity of a monotonic function $\omega:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ at $0$.

Let's rename $\omega$ by $f$, and consider the proposition that $f:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ is continuous at $0$. This is to say,

$\forall \epsilon > 0\ \exists \delta > 0\ \forall x\geq 0\ (x<\delta \Rightarrow f(x)<\epsilon)$

This can be rewritten, there exists a function $\delta:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ with $\epsilon>0\Rightarrow \delta(\epsilon)>0$ such that

$\forall \epsilon > 0\ \forall x\geq 0\ (x<\delta(\epsilon) \Rightarrow f(x)<\epsilon)$

Well this looks a lot like an adjunction doesn't it! We can make it look a bit more like an adjunction:

$\forall \epsilon > 0\ \forall x >0\ (\epsilon\leq f(x)\Rightarrow \delta(\epsilon)\leq x)$

This would be an adjunction $f\vdash\delta$, where $f,\delta:\mathbb{R}_+\to\mathbb{R}_+$, except that we have $\Rightarrow$ instead of $\Leftrightarrow$. If there was $\Leftrightarrow$, that would force $f$ not to just be continuous at $0$, but to be right-continuous throughout its domain, as discussed in this related post.

What is going on here?

Nathanael Arkor (Feb 21 2021 at 23:39):

I replied in the other thread, but this is known as a "weak adjunction". A good modern reference is Lack–Rosický's Enriched weakness.

Nathanael Arkor (Feb 21 2021 at 23:40):

A semiadjunction is something different :)

Morgan Rogers (he/him) (Feb 22 2021 at 08:41):

Joshua Meyers said:

Let $f:\mathbb{R}\to\mathbb{R}$ be a function. Consider the statement that $f$ is continuous at $x_0\in\mathbb{R}$. We can define a "modulus of continuity" $\omega:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ by $\omega(\delta)\coloneqq \text{sup}\{|f(x)-f(x_0)|:|x-x_0|< \delta\}$.

Oh hey I spent a summer thinking about this in 2016 (informally, mind you, it was while I was on holiday with my family). I had called it the $\Delta$-transform, because I was less good at coming up with names then. I'll have to dig up what I can remember about it

Morgan Rogers (he/him) (Feb 22 2021 at 09:10):

So, we can let $x_0$ vary, and let $(X,d)$ and $(Y,e)$ be arbitrary metric spaces. Then we have a transformation from the collection of functions $(X,d) \to (Y,e)$ to the collection of functions $(X,d) \times \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ sending $f$ to the mapping $(x,\delta) \mapsto \sup \{e(f(y),f(x)) : d(x,y) < \delta\}$.
We could also define $(x,\epsilon) \mapsto \sup \{d(x,y) : e(f(x),f(y)) < \epsilon \}$.
In both cases we have to allow $\infty$ as a value, to account for singularities in the first case and boundedness of $(Y,e)$ coupled with unboundedness of $(X,d)$ in the latter. The latter is actually the one I spent more time thinking about; it seems to turn failure of convexity beyond a certain threshold into failure of continuity (when $X$ is the reals, anyway), which is interesting. It also detects points at which functions are locally constant as failures of continuity in the limit $\epsilon \to 0$. I never got much further in examining the formal properties of these things or how they relate to one another, but I bet there's some good stuff in there.

Joshua Meyers (Feb 22 2021 at 15:55):

Interesting, can you say more about "seems to turn failure of convexity beyond a certain threshold into failure of continuity"?

Morgan Rogers (he/him) (Feb 22 2021 at 16:22):

Maybe convexity isn't the right description, it's more like it detects the presence of more than one turning point. For example, $x^3$ is not convex, but I don't think anything weird happens. But if you consider $x^3 - x$ with its nice curvy shape, and consider the value of $(0,\epsilon)$, it increases continuously up to $1/\sqrt{3}$ as $\epsilon \to 2/3\sqrt{3}$, where the turning points happen, and then for $\epsilon > 1/\sqrt{3}$ it jumps up to... about 1.155.
More interestingly, if I'm sketching this correctly, there is a critical value $0 < \alpha < 1/\sqrt{3}$ such that if you look at $(x,\epsilon)$ for fixed $x$ in the range $0 < x < \alpha$, then there are two points of discontinuity (and similarly below $0$), at $\pm \alpha$ there is just one point of discontinuity, and outside the range $-\alpha \leq x \leq \alpha$ there are none. So if you plot the transform, you get an $\infty$-symbol-shaped curve of points of discontinuity.

Joshua Meyers (Feb 22 2021 at 16:59):

You're talking about $(x,\epsilon) \mapsto \sup \{d(x,y) : e(f(x),f(y)) < \epsilon \}$? In my mind, for $f(x)=x^3-x$, $x=0$, this will descend continuously to $1$ as $\epsilon\to 0$ from above.

Morgan Rogers (he/him) (Feb 22 2021 at 17:39):

Yes, like I said, discontinuity of that thing as $\epsilon \to 0$ detects if a function is locally constant at $x$, which doesn't happen for any non-constant polynomial. What's interesting is the discontinuity behaviour at $\epsilon > 0$ that I was describing.

Morgan Rogers (he/him) (Feb 24 2021 at 13:48):

I also bet that there's some nice interaction between these transforms, so if anyone wants to take these ideas and run with them, I'd love to (eventually) participate in writing a paper proving some of the basic results about them.

Joshua Meyers (Feb 24 2021 at 14:29):

Morgan Rogers (he/him) said:

Yes, like I said, discontinuity of that thing as $\epsilon \to 0$ detects if a function is locally constant at $x$, which doesn't happen for any non-constant polynomial. What's interesting is the discontinuity behaviour at $\epsilon > 0$ that I was describing.

I don't see the discontinuity you are describing, can you explain? It seems to me that the function you mentioned that takes $x$ and $\epsilon$ as arguments is always continuous in $\epsilon$ for $x\in [-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}]$ and has a discontinuity in $\epsilon$ otherwise

Morgan Rogers (he/him) (Feb 24 2021 at 15:06):

It turns out my hand-drawn sketch of the function was bad. There are fewer discontinuities than I though, but there's definitely at least one for small $x$.
epsilon.png
Beyond the turning points, the function will always be growing faster in the direction of increasing $|x|$, so the location of the maximum value doesn't switch; it's that switching that allows discontinuities to happen. I thought there was one discontinuity for each turning point when $x$ is small, but only the more distant turning point actually creates a discontinuity, because once again the location of the maximum value is towards/through 0.

Here there is no discontinuity at $\epsilon = e_1$, but there is one at $\epsilon = e_2$.

Joshua Meyers (Feb 24 2021 at 15:23):

Sorry, I am confused, can you please state your claim more precisely? What function are you saying has a discontinuity where?

Morgan Rogers (he/him) (Feb 24 2021 at 16:39):

Sure, sorry. I'm saying that for $f(x) = x^3 - x$, the transformed function $(x,\epsilon) \mapsto \sup \{d(x,y) : e(f(x),f(y)) < \epsilon \}$ has a discontinuity at $\epsilon = 2/3\sqrt{3} - |f(x)|$ for $|x| < 1/\sqrt{3}$. I checked the value this time. Specifically, the value jumps from $|x| + 1/\sqrt{3}$ to $2/\sqrt{3}-|x|$.

Joshua Meyers (Feb 24 2021 at 17:05):

I see now. Let's take $x=0$ for simplicity. When $\epsilon< 2/3\sqrt{3} -|f(x)|=2/3\sqrt{3}$, the set $\{d(x,y):e(f(x),f(y))<\epsilon\}$ has 3 connected components --- I think you might be just looking at the connected component that includes $0$. For example, if $\epsilon\to 0$, this set converges to $\{0,1\}$, so the "transformed function" converges to $1$ --- does that align with your intuition?

Morgan Rogers (he/him) (Feb 24 2021 at 18:01):

Oh you're right! :relieved: What I wrote wasn't what I was actually using!
I guess what I meant was the transform $(x,\epsilon) \mapsto \sup \{\delta \mid d(x,y)< \delta \implies e(f(x),f(y)) < \epsilon \}$
... which behaves somewhat differently, but looks more like what we would be interested in with continuity as a jumping-off point.

Morgan Rogers (he/him) (Feb 24 2021 at 18:03):

and then the more natural dual transform would be
$(x,\delta) \mapsto \inf \{\epsilon \mid d(x,y) < \delta \implies e(f(x),f(y)) < \epsilon \}$

Joshua Meyers (Feb 24 2021 at 18:06):

That makes more sense. So that dual transform is identical to my function $\omega$. I'll have to think more about the other one

Morgan Rogers (he/him) (Feb 24 2021 at 18:08):

Right, they do coincide, which is how I got confused about the definition of the other one :wink: Thanks for having the patience to work out what I was doing wrong!

Joshua Meyers (Feb 24 2021 at 21:02):

So let's think about the other one: $(x,\epsilon) \mapsto \sup \{\delta \mid d(x,y)< \delta \implies e(f(x),f(y)) < \epsilon \}$

This won't be defined if that set is empty, but I guess we can define it to be $0$ then, because in $\mathbb{R}_+$, $0$ is the supremum of the empty set.

Joshua Meyers (Feb 24 2021 at 21:02):

Then we can say that $f$ is continuous at $x$ iff this function is nonzero for all values of $\epsilon$

Joshua Meyers (Feb 24 2021 at 21:05):

I guess the dual transform is a kind of measure of "efficiency": how much $\epsilon$-boundedness do you get for a given $\delta$-boundedness?

And the transform answers a similar question: how much $\delta$ boundedness do you need to get a given $\epsilon$-boundedness?

Joshua Meyers (Feb 24 2021 at 21:05):

I could see these being useful for thinking about complicated estimations

Joshua Meyers (Feb 24 2021 at 21:19):

Here's an interesting way of thinking about it: fixing $x$, define a partial order on $\mathbb{R}_+\times \{0,1\}$ by setting $(\delta,0)\leq (\delta',0)$ if $\delta\leq\delta'$, $(\epsilon,1)\leq (\epsilon'1)$ if $\epsilon\leq\epsilon'$, and $(\delta,0)\leq (\epsilon, 1)$ if for all $y$, $d(x,y)< \delta \implies e(f(x),f(y)) < \epsilon$.

Then the transforms can be written $\epsilon \mapsto \sup \{\delta \mid (\delta,0)\leq (\epsilon,1)\}$ and $\delta \mapsto \inf \{\epsilon\mid (\delta,0)\leq (\epsilon, 1)\}$

Joshua Meyers (Feb 24 2021 at 21:22):

We can also write this more cleanly by considering the injections $i_0,i_1:\mathbb{R}_+\to\mathbb{R}_+\times \{0,1\}$ defined by $i_0(\delta)=(\delta,0)$ and $i_1(\epsilon)=(\epsilon,1)$. Then we get for the transforms, $\epsilon \mapsto \sup \{\delta \mid i_0(\delta)\leq i_1(\epsilon)\}$ and $\delta \mapsto \inf \{\epsilon\mid i_0(\delta)\leq i_1(\epsilon)\}$

Joshua Meyers (Feb 24 2021 at 21:23):

And we could generalize this to any cospan of posets $A\xrightarrow{i_0}P\xleftarrow{i_1}B$ !

Joshua Meyers (Feb 24 2021 at 21:24):

This might be useful for answering questions about resource convertibility, as in resource theories

Morgan Rogers (he/him) (Feb 24 2021 at 22:10):

Joshua Meyers said:

Here's an interesting way of thinking about it: fixing $x$, define a partial order on $\mathbb{R}_+\times \{0,1\}$ by setting $(\delta,0)\leq (\delta',0)$ if $\delta\leq\delta'$, $(\epsilon,1)\leq (\epsilon'1)$ if $\epsilon\leq\epsilon'$, and $(\delta,0)\leq (\epsilon, 1)$ if for all $y$, $d(x,y)< \delta \implies e(f(x),f(y)) < \epsilon$.

Then the transforms can be written $\epsilon \mapsto \sup \{\delta \mid (\delta,0)\leq (\epsilon,1)\}$ and $\delta \mapsto \inf \{\epsilon\mid (\delta,0)\leq (\epsilon, 1)\}$

If you just consider the two copies of $\mathbb{R}_+$ separately, then that last line looks like an adjunction (cf the title of the topic)!

Joshua Meyers (Feb 24 2021 at 22:42):

It does! We can get it to look even more like an adjunction:

Let $\Omega$ be the function $\epsilon \mapsto \sup \{\delta \mid (\delta,0)\leq (\epsilon,1)\}$ and $\omega$ be the function $\delta \mapsto \inf \{\epsilon\mid (\delta,0)\leq (\epsilon, 1)\}$.

Then $\Omega(\epsilon)\leq\delta\Longleftrightarrow(\delta,0)\leq (\epsilon,1)\Longleftrightarrow \epsilon\leq\omega(\delta)$ !

Joshua Meyers (Feb 24 2021 at 22:45):

I think your two transformations form a literal adjunction!

Joshua Meyers (Feb 24 2021 at 22:51):

Now this makes me wonder, what does this say about $f$? The first thing is that $f$ is continuous at $x$ iff $\Omega$ is always non-zero, i.e. for every $\epsilon$ we can pick a non-zero $\delta$ such that etc.

Morgan Rogers (he/him) (Feb 25 2021 at 09:19):

This adjunction expresses what $x$ can "see" about $f$ using only distance information, which includes local stuff like continuity and local constant-ness, but also more subtle distant behaviour like what I worked out re the cubic. Tracing $\Omega$ out as $x$ varies for (eg) piecewise linear functions can have interesting results. I think it can also do smoothness properties like derivatives, at least for real functions.

John Baez (Feb 25 2021 at 16:09):

How could you show $f : \mathbb{R} \to \mathbb{R}$ given by

$f(x) = |x|$

is not smooth using only distance information?

$d(f(0), f(x)) = d(g(0),g(x))$

where

$g(x) = x$

is smooth, so if you want to detect the non-smoothness at $0$ you'd need to look at some point other than $0$. But actually I think

$d(f(y)),f(x)) = d(g(y),g(x))$

for all $x,y \in \mathbb{R}$. If this is true I guess it's hopeless, right?

Joshua Meyers (Feb 25 2021 at 16:48):

I don't think that's true: let $x=1,y=-1$.

Morgan Rogers (he/him) (Feb 25 2021 at 16:49):

I should really avoid making imprecise concluding statements here on Zulip, someone always ends up picking me up on it. The best that $\Omega$ can do is compute the limsup of the modulus of the derivative at $y$ as $y \to x$. If I have $f(x) = x$ for $x \geq 0$ and $f(x) = kx$ for $x < 0$ with $|k| \leq 1$, then we have $\Omega_x(\epsilon)/\epsilon \to 1$ as $\epsilon \to 0$ when $x \geq 0$ and $\Omega_x(\epsilon)/\epsilon \to |k|$ as $\epsilon \to 0$ when $x<0$.

In particular, this has the nice property of having a value even at $0$ where the function is not differentiable, and it can detect any discontinuities in $|df/dx|$ (in the case $|k|<1$, but it can't detect points at which the sign of $df/dx$ flips without changing in modulus.

Joshua Meyers (Feb 25 2021 at 16:50):

But more to the point, I think that $\omega_f(x,\delta)=\omega_g(x,\delta)$ for all $x\in\mathbb{R},\delta\geq 0$.

Morgan Rogers (he/him) (Feb 25 2021 at 16:55):

Morgan Rogers (he/him) said:

In particular, this has the nice property of having a value even at $0$ where the function is not differentiable, and it can detect any discontinuities in $|df/dx|$ (in the case $|k|<1$, but it can't detect points at which the sign of $df/dx$ flips without changing in modulus.

This can be adjusted for with some perturbations, though: add on a function with strictly positive (small) derivative everywhere; then applying $\Omega$ will detect the points where the sign suddenly changed before as discontinuities.

John Baez (Feb 25 2021 at 17:10):

I should really avoid making imprecise concluding statements here on Zulip, someone always ends up picking me up on it.

This is how math progresses! :upside_down:

There's really nothing more fun than reading a conjecture and trying to quickly disprove it with a counterexample.

Joshua Meyers (Feb 25 2021 at 17:55):

Interesting points about $\Omega_x(\epsilon)/\epsilon$. A small correction: I think you mean $\Omega_x(\epsilon)/\epsilon\to 1/|k|$, not $|k|$. So we can say $(\liminf_{\epsilon\to 0}\Omega_x(\epsilon)/\epsilon)^{-1}=\limsup_{\delta\to 0} \omega_x(\delta)/\delta = \limsup_{y\to x} |f(y)-f(x)|/|y-x|$

Joshua Meyers (Feb 25 2021 at 17:55):

So it is kind of like a derivative but it always exists

Joshua Meyers (Feb 25 2021 at 18:29):

Probably related to the Dini derivatives

Morgan Rogers (he/him) (Feb 26 2021 at 09:54):

My bad, I guess it should have been $\epsilon/\Omega_x(\epsilon)$

Morgan Rogers (he/him) (Feb 26 2021 at 09:55):

Yes, it does indeed seem like the Dini derivative, except symmetric. Perhaps $\omega$ gives the lower one?

Joshua Meyers (Feb 27 2021 at 21:43):

I think that $\limsup_{\delta\to 0} \omega_x(\delta)/\delta = \max\{D^+f(x),-D_+f(x),D^-f(x),-D_-f(x)\}$.

Joshua Meyers (Feb 27 2021 at 21:47):

Shows that $\omega$ really loses a lot of information in a "real numbers" context (rather than a general metric space context), since we're coming from both sides at once and taking the absolute value... There could be more sensitive moduli, for example:

$\omega^+(x,\delta)\coloneqq \inf\{\epsilon | (\forall y)\ 0<y-x<\delta \Rightarrow f(y)-f(x) < \epsilon\}$
$\omega_+(x,\delta)\coloneqq \inf\{\epsilon | (\forall y)\ 0<y-x<\delta \Rightarrow f(x)-f(y) < \epsilon\}$
$\omega^-(x,\delta)\coloneqq \inf\{\epsilon | (\forall y)\ 0<x-y<\delta \Rightarrow f(y)-f(x) < \epsilon\}$
$\omega_-(x,\delta)\coloneqq \inf\{\epsilon | (\forall y)\ 0<x-y<\delta \Rightarrow f(x)-f(y) < \epsilon\}$

Conjecture: these correspond to the Dini derivatives.

Ellis D. Cooper (Mar 16 2021 at 17:23):

For your information, I observed that continuity via epsilon-delta formula can be interpreted as an adjunction at the Bowdoin College Category Theory Advanced Science Seminar June 24 to August 14, 1969. Everyone was there, Eilenberg, Mac Lane, and so on. Mac Lane gave a series of lectures, and kindly presented my idea to the (large) audience of categorists. Eilenberg wrote for me a letter of recommendation to attend. I had the good fortune to have been seated next to the analyst, William F. Donoghue, Jr., and I had told him my idea, but was stuck on a detail. He told me about the modulus of continuity, and so began (and ended) my fame as a categorist.