# Continuous maps between $\mathbb N\cup \{\infty\}$ and $\mathbb R$

Consider the set $$N=\mathbb N\cup\{\infty\}$$ together with the following topology: a subset $$U$$ of $$N$$ is open if either $$\infty\notin U$$ or $$N\setminus U$$ is finite.

(1) Describe continuous maps $$\mathbb R\to N$$ and $$N\to \mathbb R$$.

(2) Does there exist a subset of $$\mathbb R$$ homeomorphic to $$N$$?

--

(1) I'm not quite sure what is being asked. A continuous map is one with the property that preimages of open sets are open. We know how open sets look like in both spaces. But what exactly can I conclude about continuous maps?

(2) It looks like $$N$$ is compact. So the only candidates for such subsets are compact subsets of $$\mathbb R$$. But I guess I need to understand (1) first? If I do, I will have understand how restrictions of continuous maps look like as well, I suppose.

• Recall that $\mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $\mathbb{R} \to N$? For the other direction, can you extend a function $\mathbb{N} \to \mathbb{R}$ to a continuous map $N \to \mathbb{R}$? If no, what condition do you need? – Luca Carai Dec 3 '18 at 1:57
• @LucaCarai So from what you said we can conclude that if $\mathbb R\to N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions. – user531587 Dec 3 '18 at 2:19
• Go back to the definition of connected sets. Let U = $U_1 \sqcup U_2$. What does the topology you have say about U? – Joel Pereira Dec 3 '18 at 2:28
• @JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $\infty$, or they both have finite complements, or one of the does not contain $\infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open... – user531587 Dec 3 '18 at 3:08

(1) a continuous map $$f$$ from $$\mathbb{N} \cup \{\infty\}$$ into $$\mathbb{R}$$ corresponds to a convergent sequence and its limit, in the sense that for any space $$X$$, $$f: \mathbb{N} \cup \{\infty\} \to X$$ is continuous iff $$x_n = f(n)$$ defines a sequence that converges to $$f(\infty)$$ in $$X$$. And conversely for every sequence $$x_n$$ in $$X$$ that converges to $$x$$, the function defined by $$f(n) = x_n$$ for all $$n$$ and $$f(\infty) = x$$, is continuous from $$\mathbb{N} \cup \{\infty\}$$ to $$X$$.
The other way around (from $$\mathbb{R}$$ to $$\mathbb{N} \cup \{\infty\}$$ there are only constant maps as $$\mathbb{N} \cup \{\infty\}$$ is totally disconnected, and $$\mathbb{R}$$ is connected and thus has connected image.
Any convergent sequence with limit (like $$\{\frac{1}{n}: n \ge 1\} \cup \{0\}$$) is homeomorphic to $$\mathbb{N} \cup \{\infty\}$$, as is easily checked.
• I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(\infty)$, we need to show that any neighborhood of $f(\infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(\infty)$ there exists a nbhd of $\infty$ whose image lies in the nbhd of $f(\infty)$. Any nbhd of $\infty$ contains infinitely many elts of $N$, so the nbhd of $f(\infty)$ also contains infinitely many pts. Is that how the proof goes? – user531587 Dec 5 '18 at 3:55