For each $\epsilon$, for which $\delta$ does $d(x,y)<\delta\implies d(f(x),f(y))<\epsilon$ hold?

Question

I have the definition : A function from a metric space to a metric space is uniformly continuous if for all $\epsilon>0$ there exists $\delta$ such that $d(x,y)<\delta\implies d(f(x),f(y))<\epsilon$.

Therefore for each $\epsilon$ we can identify some unique largest $\delta$ for which the implication always holds. What are the values of $\underline\delta=\lim \sup \delta(\epsilon)$ for each corresponding $\epsilon$ in the following case?

Let $X=\Bbb Z[\frac16], d$ where $d(x,y)=\lvert x-y\rvert_2$, the 2-adic metric.

Then let $f:X\to X$

$f(x)=\begin{cases}3x+2^{\nu_2(x)}\cdot3^{\nu_3(x)}&x>0\\f(x)=0&x=0\\3x-2^{\nu_2(x)}\cdot3^{\nu_3(x)}&x<0\end{cases}$

Now I believe I've determined $f$ is continuous and since $X$ inherits its topology from a compact space that should make $f$ uniformly continuous. So I want to verify this either algebraically or with some values and determine which delta and epsilon correspond.

But I'm not getting results that obviously confirm the proposition. For example , for pairs $(x,y)$ I get the following results for the corresponding ceiling/floor on delta and epsilon:

$(5,85): \underline\delta(\frac1{16})=\frac18$

$(13,53): \underline\delta(\frac18)=\frac18$

$(21,85): \underline\delta(\frac12)=\frac1{64}$

So I'm beginning to doubt the premise that $f$ is continuous. What is the $\delta$ for each $\epsilon$, or else where did I going wrong in understanding and determining continuity of $f$?

---

Here's my argument of continuity of $f$:

Firstly, by reflectivity of both the function and the metric around $0$ we can ignore negative values and prove for non-negative values without loss of generality.

As a function from the odd numbers onto the odd numbers $f(x)=(3x+1)\lvert3x+1\rvert_2$ is discontinuous only at $-\frac13$ where $f(-\frac13)=0$

Then extending the same map to any isobar of $\Bbb Z[\frac12]$ by perserving its powers of $2$ as follows: $f(2^nx)=2^nf(x)$ is only discontinuous at $f(-\frac{2^m}{3})=0:m\in\Bbb Z$

Then modifying the map to eliminate the leaves of its graph, and at the same time eliminate maps into $0$ we have $f(x)=3x+2^{\nu_2(x)}\cdot3^{\nu_3(x)}$ which should be continuous. However I'm unsure, perhaps the continuous map is:

$f(x)=(3x+2^{\nu_2(x)}\cdot3^{\nu_3(x)})\lvert3x+2^{\nu_2(x)}\cdot3^{\nu_3(x)}\rvert_2$

Answer 1

What is $\mathbb{Z}[\frac{1}{6}]$ and is $\nu_2$(x)? I'm not familiar with that notation. Also, if a function is uniformly continuous, the $\delta$ only depends on $\epsilon$. I suspect that since $\mathbb{Z}$ is discrete, you can pick $\delta$ < 1, but that depends on the definition of X.

Answer 2

For each $\epsilon$, for which $\delta$ does $d(x,y)<\delta\implies d(f(x),f(y))<\epsilon$?

According to this, if you choose a particular pair $(x,y)$ and calculate $d(x,y)$ and $d(f(x),f(y)),$ $\delta$ must be greater than $d(x,y)$ and $\epsilon$ must be greater than $d(f(x),f(y)).$ How do you get values to which $\delta$ and $\epsilon$ must be equal?

In general there is no reason to expect that $d(x,y)$ will be a function of $d(f(x),f(y)),$ let alone a one-to-one function. The mere fact that you have three points that don’t fit whatever pattern you expected them to fit is meaningless.

——

Update: Since the question has changed, I’ll address some of the new points raised.

Therefore for each $\epsilon$ we can identify some unique $\delta$ for which the implication always holds.

That is true if you delete the word “unique.” Consider $\mathbb R$ with the usual metric, and consider a constant function $\mathbb R\to\mathbb R.$ The function is uniformly continuous, but any positive $\delta$ suffices. What makes any value of delta “unique”?

What are the values of $\underline\delta=\lim \sup \delta(\epsilon)$ for each corresponding $\epsilon$ in the following case?

In the example I just gave, you could take the lim inf (not lim sup) of $\delta,$ which would be $0$ for any positive $\epsilon.$ Note that the lim inf is not itself a value of $\delta$ that satisfies the condition.

You cannot find the lim inf (or lim sup) of $\delta$ for any particular $\epsilon$ merely by looking at one pair $(x,y).$

Also note that the same comment that suggested there might be a minimum or maximum $\delta$ also gave the opinion that this result would be useless.

In fact, what your three examples of $\underline\delta(\epsilon)$ appear to show is that your treatment of $\underline\delta$ is incorrect.