# Let $(Y,d')$ be another metric space, when is $f:X\rightarrow Y$ continuous?

I'm having some trouble with the last question asked to me in an assignment

I have an initial metric space given as:

$$d:X\times X \longrightarrow \mathbb{R}_{\geq0}$$, $$d(x,y):= \begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x\neq y % \end{cases}$$

I've identified that $$(X,d)$$ is indeed a metric space, and that all of the positive whole numbers in $$X$$ are open sets. My question is for another metric space $$(Y,d')$$, when will a function $$f:X\rightarrow Y$$ be continuous?

My hypothesis is that I need to use an $$\varepsilon$$, $$\delta$$ proof where I select a delta sufficiently large to prove that for every $$\varepsilon >0$$ there exists a $$\delta>0$$ so that for all $$\hat{x}\in X$$ with $$d(x,\hat{x})< \delta$$ we have $$d'(f(x),f(\hat{x}))<\varepsilon$$. However where I'm stuck is how to prove this in a proper way, as I haven't been able to find comparable examples.

Every function is continuous then. Given $$\varepsilon>0$$, you take $$\delta=1$$ and then\begin{align}d(x,y)<\delta&\iff d(x,y)<1\\&\iff x=y\\&\implies d'\bigl(f(x),f(y)\bigr)=0<\varepsilon.\end{align}
• Hi José - does this imply that $f$ will be continuous for every $x\in X$ ? I think it seems to be the case – chactas Nov 21 '19 at 12:35
• I don't really understand your question. I wrote that $f$ is continuous then. That means that $f$ is continuous at every point, right?! By the way, what I did proves that $f$ is actually uniformly continuous. – José Carlos Santos Nov 21 '19 at 12:37