The following question reads as
i) Given an example of infinite metric spaces $(X,d)$ and $(Y,\delta)$ such that every function $f\colon X\to Y$ is continuous.
ii) Is it possible to give an example where no functions $f\colon X \to Y$ are contiuous? (Or very few???)
I originally thought to have an understanding on sets and their coupled metrics. But now I am struggling to imagine what is going on here.
For question (i) I realise that I want to show that every function, $f:X \to Y$ is continuous INDEPENDENT of what $f$ actually is. Therefore, given any open set $U$ in $Y$, I am required to show that $f^{-1}(U)=V$ is always be open. Given this, I think that the metric on $Y$ is irrelevant to the open-ness of $X$. If I assign the discrete metric on $X$, I see that no matter what $f^{-1}(U)$ yields, I should always be able to have a ball of radius $\frac{1}{2}$ around each $x \in f^{-1}(U)$. Thus $f^{-1}(U)$ is open. In addition, the infinite sets I pick, can they be anything since they'll always be separated by 1? Does $\mathbb{R}$ suffice?
For question (ii) I realise that I must possibly find the opposite. That is, find the metric that will have every $f^{-1}(U)$ closed. Does $d(x,y)=0$ for all $x,y \in X$ work? I am also a little confused by the use of plurals in the question. "Possible to give EXAMPLES", "or very few"? I am assuming there is something going on that I am missing.
Thank you in advanced for you help.
For your second example (with only few continuous functions), let $X$ be $\mathbb{R}^n$ with the euclidean metric and $Y$ be $\mathbb{R}^n$ with the discrete metric. Then the only continuous functions are the constant ones. To see this, let $y\in Y$. The set $\{y\}$ is open and closed, so for a continuous function $f:X\rightarrow Y$, the set $f^{-1}(\{y\})$ is open and closed. But in euclidean space, the only sets that are open and closed are the empty set and the whole space. So $f$ must be constant.