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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 $N$?

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(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.

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  • $\begingroup$ 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? $\endgroup$ – Luca Carai Dec 3 '18 at 1:57
  • $\begingroup$ @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. $\endgroup$ – user531587 Dec 3 '18 at 2:19
  • $\begingroup$ Go back to the definition of connected sets. Let U = $U_1 \sqcup U_2$. What does the topology you have say about U? $\endgroup$ – Joel Pereira Dec 3 '18 at 2:28
  • $\begingroup$ @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... $\endgroup$ – user531587 Dec 3 '18 at 3:08
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(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.

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  • $\begingroup$ In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement? $\endgroup$ – user531587 Dec 5 '18 at 2:18
  • $\begingroup$ For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/… $\endgroup$ – user531587 Dec 5 '18 at 2:34
  • $\begingroup$ @user531587 I made the statement exact about sequences. $\endgroup$ – Henno Brandsma Dec 5 '18 at 2:52
  • $\begingroup$ 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? $\endgroup$ – user531587 Dec 5 '18 at 3:55
  • $\begingroup$ @user531587 not just infinitely many; all but finitely many which is stronger and gives convergence. $\endgroup$ – Henno Brandsma Dec 5 '18 at 3:57

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