# How can I prove that any function $f: \mathbb{N} \rightarrow \mathbb{R}$ is continuous?

I have to find all functions of the type:

$$f: \mathbb{N} \rightarrow \mathbb{R}$$

that are continuous. My claim is that all such functions are continuous. If we think about it, $$f$$ would only have isolated points and we know that a function is always considered continuous at an isolated point. So, by this reasoning, any function $$f:\mathbb{N} \rightarrow \mathbb{R}$$ is continuous. I hope my reasoning is correct. What I am confused about is how could I prove this formally. We know that continuity is defined like this:

A function $$f: A \rightarrow \mathbb{R}$$ is continuous at a point $$c \in A$$ if

$$\forall V \in \mathcal{V}(f(c)), \exists U \in \mathcal{V}(c) \text{ such that } \forall x \in U \cap A \text{ we have } f(x) \in V$$

How could I possibly prove my point using the definition of continuity? It's really not that difficult to find the answer intuitively, but I don't see how I could make my argument more formal.

• Bro... isolated points are never continuous...
– user838035
Commented Nov 7, 2020 at 14:58
• You need a topological definition of continuity. Then if $\mathbb N$ has the discrete topology, any function from $\mathbb N$ is continuous. Commented Nov 7, 2020 at 15:00
• @SenZen You are incorrect. You have to respect the topology of $\mathbb{N}$, which is the discrete one. So every subset of $\mathbb{N}$ is open. A function is continuous if the preimage of an open set $U\subseteq \mathbb{R}$ is open. But as every subset is open with regards to the discrete topology we have that $f^{-1}(U)\subseteq\mathbb{N}$ is always open. Hence $f$ is continuous. Commented Nov 7, 2020 at 15:00
• Wait what on earth is that definition of continuity? I know of the epsilon-delta definition, but I've never seen that one, and it makes no sense to me - what is $\mathcal{V}$??
– user838035
Commented Nov 7, 2020 at 15:01
• Ah I think this should be tagged under topology then, not real analysis...
– user838035
Commented Nov 7, 2020 at 15:06

For each $$c\in\Bbb N$$, take $$\mathcal V(c)=\{c\}$$. You can do it, since it is a neighborhood of $$c$$.

• But I don't think I can do that. At least the way I was taught, a neighborhood of $x \in \mathbb{R}$ is an interval of the type $(x - \epsilon, x + \epsilon)$ for some $\epsilon > 0$. We were told that $\epsilon$ can be arbitrarily small, but it can't really be $0$. So what should I do about that?
– user719014
Commented Nov 7, 2020 at 21:06
• I am sorry, but that is not the definition of neighborhood. Or are you claiming that $(-1,2)$ is not a neighborhood of $0$? Commented Nov 7, 2020 at 21:08
• Yes, it is a neighborhood according to that definition, but what you wrote was that a neighborhood of $x$ is an interval of the type $(x-\varepsilon,x+\varepsilon)$, and $(-1,2)$ is not an interval of the form $(0-\varepsilon,0+\varepsilon)$. Commented Nov 7, 2020 at 21:19
• Anyway, a neighborhood of a natural number $c$ in $\Bbb N$ is a set $V$ that contains $(c-\varepsilon,c+\varepsilon)\cap\Bbb N$, for some $\varepsilon>0$. So, $\{c\}$ is a neighborhood of $c$ in $\Bbb N$, since it is equal to $(c-1,c+1)\cap\Bbb N$. Commented Nov 7, 2020 at 21:21
• Got it. Thanks for putting up with me :)
– user719014
Commented Nov 7, 2020 at 21:23