Supposing you have uniformly continuous function between metric space, is the epsilon delta relationship continuous. More precisely, for $X,Y$ metric, given $f :X \rightarrow Y$, uniformly continuous, if for a given $\varepsilon>0$ we define the $\delta(\varepsilon)$ to be the sup of all $\delta>0$ s.t $ d_X(x,y)< \delta \Rightarrow d_Y(f(x),f(y))< \varepsilon $ , is  $g:\mathbb{R}^+ \rightarrow\mathbb{R}^+$  given by $g(\varepsilon)= \sup${ $\delta$ >0 that work }, a continuous a function?
Edit: It occurs to me that the sup defined here way me infinity, as for constant functions. If we exclude these cases, is the above true?
 A: Consider a sawtooth wave from the reals to the reals...but let's make the rising slope be 1, and the falling slope be 2. That's uniformly continuous, with $\delta = \epsilon/2$ working everywhere. But on the rising segments, your $\delta_{max}$ will be $1$, and on the falling segments, it'll be $1/2$, and hence it must be discontinuous at each "breakpoint". Actully, the same argument works for
$$
f(x) = \begin{cases}
x & -1 \le x \le 0 \\
-2x & - \le x \le \frac12
\end{cases}
$$
which is a uniformly continuous function from $[-1, 0.5]$ to $[-1, 0]$.
A: It does not necessarily have to. Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x+2n) = n$, $f(x+2n+1) = n+x$ for $x \in [0, 1)$ and $n \in \mathbb{Z}$. (The graph of this function shall look like a staircase) Note that $f$ is $1$-Lipschitz so is uniformly continuous.
We observe that $\delta(\epsilon) \ge 2$ for $\epsilon > 1$. Indeed, if $x<y$ satisfy $f(y) - f(x) \ge \epsilon$, then $m((x, y) \cap \bigcup_{n \in \mathbb{Z}} (2n+1, 2n+2))$ should be bigger than $\epsilon > 1$. Since intervals $(2n+1, 2n+2)$ are of size 1 and are separated from each other by distance $\ge 1$, we must have $|y-x| \ge 2$.
However, $\delta(1) = 1$. Indeed, $|f(2) - f(1)| = 1$ prohibits us from taking $\delta$ greater than 1 for $\epsilon = 1$. In contrast, any $\delta$ smaller than 1 will work. Thus we have a discontinuity of $\delta(\epsilon)$ at $\epsilon=1$.
