I've just started topology on my grade. It makes a lot of sense to start introducing something that we've already been introduced to before on Calculus I, the notion of metric spaces.
My teacher said in class that if we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider.
"After that, we find the motivation to think that distance is not important at all, what's really important is to find the open sets, so welcome to topology." Quoting her.
So I don't see as clear as her this sentence: "If we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider."
Can someone provide an intuitive explanation or a hint to deduce a formal proof to that?
Thanks for your time.
Best Answer
I cannot explain that since it is trivially false. For instance, if we, on $\mathbb R$, consider the discrete distance$$d(x,y)=\begin{cases}1&\text{ if }x\neq y\\0&\text{ otherwise.}\end{cases}$$ and if $d'$ is the usual distance, then $\{0\}$ is an open set in $(\mathbb{R},d)$ (being equal to $B_1(0)$), but not in $(\mathbb{R},d')$.