Does there exist topology $T$ on the real numbers such that the class of $T$-$T$-continuous and uniformly continuous functions coincide?

Question

Does there exists a topology $T$ on the set of real numbers $\mathbb{R}$ such that the $T$-$T$-continuous functions from $\mathbb{R}$ to $\mathbb{R}$ are precisely the uniformly continuous functions? By uniformly continuous, I mean the standard definition of uniformly continuous.

Answer 1

Such topology doesn't exist, this is just a simple modification of answer by RT1 in another post.

Suppose there exists a topology $T$ on $\mathbb{R}$ such that $f:\mathbb{R}\to\mathbb{R}$ is uniformly continuous in Euclidean metric iff $f:\mathbb{R}\to\mathbb{R}$ is $T$-$T$-continuous.

1. $T$ is $T_1$: See argument in this answer by RT1. Let $a, b\in\mathbb{R}$ be distinct, pick non-empty proper open set $B$ (such must exists since $T$ must be non-trivial) and let $A$ be its complement, let $f$ map $A$ to $\{a\}$ and $B$ to $\{b\}$. Then since $f$ is not uniformly continuous, its not $T$-$T$-continuous, so $A$ is not open. There is $U\in T$ with $f^{-1}(U) = A$ and then $U$ is a neighbourhood of $a$ which is not a neighbourhood of $B$.

2. Closed rays are closed with respect to $T$. Let $[a, \infty)$ be an interval and $f$ be defined to be constant $c$ on $[a, \infty)$, and linear and injective on $(-\infty, a]$. Since $f$ is uniformly continuous and $T$ is $T_1$, we have that $f^{-1}(c) = [a, \infty)$ is closed. Similar argument will show that $(-\infty, a]$ is closed.

3. $T$ contains the Euclidean topology. Since open rays are open, all open intervals are open, so all open sets in Euclidean topology are open.

Then its simple, let $f(x) = x^2$, then $f$ restricted to any bounded open interval $(a, b)$ has a uniformly continuous extension (for exaple, extend it to be linear on $(-\infty, a]$ and $[b, \infty)$), so $$f\restriction_{(a, b)}:((a, b), T\restriction_{(a, b)})\to (\mathbb{R}, T)$$ is continuous. From gluing lemma it follows that $f$ is $T$-$T$-continuous, but this is impossible since $x\mapsto x^2$ is not uniformly continuous.

In fact all of the uniformly continuous functions considered above are Lipschitz, so this also proves that there is no such topology for Lipschitz continuous functions.