Showing that a function is not $(d,d)-$ continuous at a point.

Question

Let $d: \mathbb R \times \mathbb R \rightarrow \mathbb R$ be a metric: $$ d(x,y) = \begin{cases} 0 & x = y \\ |x| + |y| + 3|x-y| & x \neq y \end{cases} $$ Show that the function $f: \mathbb R \rightarrow \mathbb R$ given by $f(x) = (1+x)^2$ is not $(d,d)-$ continuous at $c = 0$.

So we know that the image of $x=0$ is $f(0)=1$. Now, for any fixed $\epsilon>0$, we need to show that there does not exist such a $\delta$ that $f(B_\delta(c))\subseteq B_\epsilon(f(c))$, right? How do we proceed with that?

Answer 1

No, find one fixed $\varepsilon$, that you can choose and show that for every $\delta > 0$ we can find a point $x_\delta$ with $d(x_\delta,0) < \delta$ and $d(f(x_\delta),f(0)) \ge \varepsilon$.

The latter implies indeed that $f[B_\delta(0)] \nsubseteq B_\varepsilon(f(0))$.

Note that $d(x,1) \ge 1$ for any $x \neq 1$. Also note that near $0$ we have $d(0,x) = 4|x|$...