Implications of continuity

Question

A mapping $T:X\rightarrow Y$ between metric spaces $(X.d),(Y,\hat d)$ is continuous at $x_0$ if for every $\epsilon>0$ there is a $\delta>0$ such that $\hat d(Tx,Tx_0)<\epsilon$ for all $x$ satisfying $d(x,x_0)<\delta$.

Consider such a mapping $T$, continuous at $x_0$. For any $\epsilon>0$ define $$\delta(\epsilon)=\sup\{d(x_0,y): \hat d(Tx_0,Ty)<\epsilon\}.$$ Does $\epsilon_1<\epsilon_2$ imply $\delta(\epsilon_1)<\delta(\epsilon_2)$?

It seems obvious that $\delta(\epsilon_1)\leq\delta(\epsilon_2)$. Otherwise, for some $y\in X$ within $\delta(\epsilon_1)$ from $x_0$ it would be $\hat d(Tx_0,Ty)\geq\epsilon_2>\epsilon_1$. Is the inequality strict?

It's easy to see it's not strict in general, just take $\hat d$ to be some discrete metric, or something of the sort — Jakobian, Feb 16 '19 at 11:24
That's right. Do you happen to know if there is a class of mappings for which it holds, or have some hints about where to look deeper into the matter? — cangrejo, Feb 16 '19 at 11:29

score 1 · Accepted Answer · answered Feb 16 '19 at 11:39

As pointed out by Jakobian in the comments, a counterexample can easily be found.

Consider the discrete metric $d(x,y)=0$ if $x=y$, 1 otherwise, and the identity mapping from $(X,d)$ onto itself. For any $\epsilon_1, \epsilon_2 \in (0,1)$, $\delta(\epsilon_1)=\delta(\epsilon_2)$.

Implications of continuity

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