# Metrics and Continuous Functions

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.

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For 1) to happen you need discrete topology (every subset of $X$ is an open set) on $X$. 2) Second one is impossible, as you can always construct a cts function. – DiffeoR Mar 27 '14 at 10:00
But I don't know what metric will induce discrete topology on $X$. – DiffeoR Mar 27 '14 at 10:04
Yes, I forgot to mention the first answer holds when Y has any topology. – DiffeoR Mar 27 '14 at 10:06
Topology means : you precisely say which sets you call open sets in $X$. And these open sets satisfy property like countable union of them and finite intersection of them is open again. When you say a metric space : it automatically determines which sets will be called open. You first say all open balls of radius 'r' for all r > 0 are the open sets first and then apply finite intersections and countable unions of them to obtain the other open sets. So a metric determines a topology but the other way round is not true always. – DiffeoR Mar 27 '14 at 10:11
The cts function in the second case I was talking about is the constant function, i.e $f(X) = \{y_0\}$. – DiffeoR Mar 27 '14 at 10:15

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.
@eXtremiity Yes, the constant function is always continuous. To see this, let $f$ be constant and $y$ the value it takes. Let $U\subset Y$ be open. If $y\in U$, then $f^{-1}(U)=X$, which is open. If $y\not\in U$, then $f^{-1}(U)=\emptyset$, which is also open. – Vincent Boelens Mar 27 '14 at 14:39