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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.

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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}

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  • $\begingroup$ Hi José - does this imply that $f$ will be continuous for every $x\in X$ ? I think it seems to be the case $\endgroup$ – chactas Nov 21 '19 at 12:35
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Nov 21 '19 at 12:37
  • $\begingroup$ Hi José, understood, thank you. $\endgroup$ – chactas Nov 21 '19 at 12:38

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