# Describing continuous functions

I wanna describe all the continuous functions from set $\mathbb{N}$ with metric $d(x,y) = |\dfrac{{1}}x - \dfrac{{1}}y|$ to $\mathbb{R}$ with the usual metric (|x-y|). I've done proving a function is continuous before, but not describing. Here's my line of thought: let f be an arbitrary continuous function at x $\in N$, $\forall \epsilon>0, \exists \delta>0$ so that if $d_\mathbb{N}(x,y) = |1/x - 1/y| < \delta$ then $d_\mathbb{R}(f(x),f(y)) = |f(x) - f(y)| < \epsilon$. I'm stuck.

-In another way, is it true to say the metric in N acts like a discrete space and hence, any function is continuous?

• What are m and n? Shouldn't this be x and y?
– user370967
Feb 20 '18 at 22:49
• mistyping. edited
– Jack
Feb 20 '18 at 22:51
• What happens if your argument tends to infinity? One may think of $\mathbb{N}$ equipped with your metric as of a subset of $\{\frac{1}{n} \; : \; n\in \mathbb{N} \} \subset \mathbb{R}$ with standard euclidean metric. Feb 20 '18 at 22:52
• If you are given any $x\in\mathbb{N}$, there is a $\delta$ such that the only element $y\in \mathbb{N}$ with $d(x,y)<\delta$ is $x$ itself. From this follows that every function (no matter where to) is continuous at $x$. Feb 20 '18 at 22:53
• Not every function from a netric space into a metric space is continuous. For example there are plenty of non-contunuous functions from and to the real line (usual metric). Feb 20 '18 at 22:57

A metric space $X$ is discrete if for every point $x\in X$ there is a $\delta_x$ such that $\{y\in X\mid d_X(y,x)<\delta_x\}=\{x\}$.
Observe that every function defined on a discrete metric space is continuous: for every point $x$ and every $\varepsilon>0$, you can choose $\delta=\delta_x$ as above. Then if $d_X(x,y)<\delta_x$ then $x=y$ and hence $|f(x)-f(y)|=0<\varepsilon$.
Now we only need to check that $\mathbb N$ with the the metric you provided is discrete: given $x$, you can choose $\delta_x:= \frac 1{(x+1)x}$. Indeed if $x\leq y$ and $\frac 1x-\frac 1y< \frac 1{(x+1)x}$ then $(x+1)y-(x+1)x< y$, i.e. $x+1> y$, therefore $x=y$. Similarly if $x\geq y$ and $\frac 1y - \frac 1x<\frac 1{(x+1)x}<\frac1{(x-1)x}$ then $(x-1)x-(x-1)y<y$, i.e. $(x-1)<y$; therefore $x=y$.