A continuous function that isn't locally uniformly continuous

Question

For a function $f:X\to Y$ in which $X$ and $Y$ are metric spaces i say that $f$ is locally uniformly continuous if $\forall x\in X, \exists I_x$ a nbd of $x$ s.t.$f|I_x$ (the restriction) is uniformly continuous. I was wandering for a continuous function that isn't locally uniformly continuous, but i searched in Mathstackexchange but i found only this, the problem is that this function has a problem only in one point i would see a continuous function s.t. no point of his domain has a nbd in which the function is uniformly continuous (we can call this a continuous strongly not locally uniformly continuous function). Obviously, by Cantor's theorem the domain cannot have point with compact nbds.

Answer 1

On $\mathbb R,$ let $f(x)= 0, x<0,  f(x) = 1, x\ge 0.$ Let $q_1,q_2, \dots$ be the rationals. For $x\in \mathbb R,$ define

$$g(x) = \sum_{n=1}^{\infty}\frac{f(x-q_n)}{2^n}.$$

The sum converges uniformly on $\mathbb R,$ and the $n$th summand is continuous everywhere except at $q_n,$ where it jumps by $1/2^n.$ Thus $g$ is continuous at each irrational.

Let $X$ be the set of irrationals, $Y=\mathbb R.$ We then see $g|_X: X\to Y$ is continuous. Let $x_0\in X$ and let $r>0.$ Claim: $g$ is not uniformly continuous on $(x_0-r,x_0+r)\cap X.$ Proof: $(x_0-r,x_0+r)$ contains a rational $q_m.$ I think you can see what is going to happen: At $q_m,$ $g$ jumps by $1/2^m.$ That will remain true when we restrict $g$ to $X.$ Thus there will be irrationals $x_1,x_2,$ arbitrarily close to $x_0,$ such that $g(x_2)-g(x_1) > 1/2^m.$ It follows that $g|_X$ is not uniformly continuous on $(x_0-r,x_0+r)\cap X.$