Let $(X,d)$ be a metric space. Let $\overline{d}(u,v)=\min\{d(u,v),1\}$. Then are $d$ and $\overline{d}$ topologically equivalent.
I have a specific question on the following 'proof' of this statement.
First it is shown, that $\overline{d}$ is indeed a metric. That is easy. Then they show, that every $\overline{d}$-open set is $d$-open. That is easy, too.
When it is shown, that every $d$-open set is $\overline{d}$-open it goes as follows:
Let $U\subseteq X$ be $d$-open and $u\in U$. Then exists $\varepsilon > 0$ such that $B_\varepsilon^d(u)\subseteq\color{red}{X}$. Let $\delta=\min\{\varepsilon, \frac12\}\Rightarrow B_\delta^d(u)\subseteq\color{red}{X}$ and $B_\delta^d(u)\color{red}{=}B_\delta^\overline{d}(u)\Rightarrow$ U is $\overline{d}$-open.
I marked my problems red. Why is it said, two times, that we have a subset of $X$, which is trivial. Should we not say, that we have subsets of $U$? I mean, it is not wrong, but it seems odd.
Also I do not see, why we have equality, which should also not be needed to state, that $U$ is $\overline{d}$-open. Is it not enough to show, that $B_\delta^d(u)\subseteq B_\delta^\overline{d}(u)$?
Am I mistaken?
Thanks in advance.
The two red instances $X$ should indeed be $U$, the set we're trying to show openness of.
As we have $\delta=\min({\epsilon,\frac12})$ so $\delta \le \frac12 < 1$, we have $B^d_\delta(x)=B^{\bar d}_\delta(x)$: $y \in B^d_\delta(x)$ iff $d(x,y) < \delta$ iff $\bar{d}(x,y) (= \min(d(x,y),1)=d(x,y)) < \delta$ iff $y \in B^{\bar d}_\delta(x)$.