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Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{N}$ and all continuous functions $f: \mathbb{N} \rightarrow \mathbb{R}$.

My thinking process went something like this. For the case of $f: \mathbb{R} \rightarrow \mathbb{N}$, if I think about the function in the $xOy$ plane, if we would have any point at which the value would change from one natural number to some other natural number then at that point we would have a jump discontinuity. So every number from the domain $\mathbb{R}$ needs to be mapped to the same natural number in order to have a continuous function. Thus, we need the function to be something like

$$f:\mathbb{R} \rightarrow \mathbb{N} \hspace{1cm} f(x) = n$$

for any $n \in \mathbb{N}$.

In the case of $f: \mathbb{N} \rightarrow \mathbb{R}$ again thinking about the function in the plane $xOy$, the values of the function at two consecutive points $n$ and $n+1$ are not 'tied' together by anything, there's just empty space, so the function is nowhere continuous. Thus, there are no continuous functions $f: \mathbb{N} \rightarrow \mathbb{R}$.

I hope my reasoning is correct. But my real problem is about the writing process of this proof. Obviously I can't write on the paper all of this story that I just came up with. But how can I create a rigorous proof with what I just wrote (with definitions and all of that fluff). Thinking in terms of pictures is nice, but I have to formalize my thinking with definitions, theorems, and the like and in that regard I am lacking terribly. So how can I approach the writing of this proof?

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    $\begingroup$ If you want to speak about continuity, you need some topology on both spaces. On $\Bbb R$ is quite clear, but... What topology are you considering over $\Bbb N$? $\endgroup$ Nov 4, 2020 at 19:39
  • $\begingroup$ You also need to have a formal definition of continuity to work from. The graph of a function being a single connected curve is possible, but it is not the conventional choice. $\endgroup$
    – Arthur
    Nov 4, 2020 at 19:42
  • $\begingroup$ For second case constant functions like in first case obbiously also work, regardless of topology. $\endgroup$
    – zwim
    Nov 4, 2020 at 19:46
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    $\begingroup$ Since this is an Introduction to Analysis class, I think it is most likely that the topology on these spaces are the standard topology on $\mathbb{R}$ and the subspace topology on $\mathbb{N}$ @TitoEliatron $\endgroup$ Nov 4, 2020 at 19:54
  • $\begingroup$ @TitoEliatron Unfortunately I am not familiar with the term of 'topology', we haven't studied anything related to it so far. I'm sorry, but I can't clarify this. $\endgroup$
    – user592938
    Nov 4, 2020 at 23:54

1 Answer 1

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Since this is an Introduction to Analysis class, I think it is most likely that the topology on these spaces are the standard topology on $\mathbb{R}$ and the subspace topology on $\mathbb{N}$. The subspace topology on $\mathbb{N}$ is equivalent to the discrete topology on $\mathbb{N}$, so every function $f:\mathbb{N} \to \mathbb{R}$ is continuous. On the other hand, the inclusion $i:\mathbb{N} \hookrightarrow \mathbb{R}$ is continuous. So then if we have a continuous function $f:\mathbb{R} \to \mathbb{N}$, the composition $i \circ f$ is continuous. But then, if $| \text{im}f | \neq 1$ (i.e. if $f$ does not collapse $\mathbb{R}$ to a single point), then we have a continuous function $i \circ f$ that sends the connected space $\mathbb{R}$ to a disconnected subset. But if $g: X \to Y$ is a continuous function, $X$ is connected if and only if $\text{im} g$ is connected. So the only functions $f: \mathbb{R} \to \mathbb{N}$ that are continuous are those that map the reals to a singleton.

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