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.
(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.