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Let $\mathbb{N}$ have the co-finite topology, and let $\mathbb{R}$ have the usual topology. Then what functions from $\mathbb{N}$ to $\mathbb{R}$ are continuous?

I think the constant functions would be continuous. The range of any such function (not constant) has to be countable. One can pick an open interval around one of $f(x')$, $x' \in \mathbb{N}$, such that no other $f(x)$ for $x \in \mathbb{N}$ lies in that open interval. The inverse of this open set under $f$ would not be open in co-finite topology being a finite set. Thus constant functions are the only continuous functions.

I am not sure if my argument is correct.

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    $\begingroup$ The implication that if range of $f$ is countable then one can pick an open interval around one of $f(x')$, $x' \in \mathbb{N}$, such that no other $f(x)$ for $x \in \mathbb{N}$ lies in that open interval is faulty. What if the range is $\mathbb{Q}$ for example? $\endgroup$ – Seven Aug 8 '15 at 8:53
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Your argument doesn't quite work, because the range could be a finite subset of $\mathbb R$ containing at least two elements. Then there is no guarantee that $f^{-1} (y)$ of the particular $y = f(x')$ that you have chosen is not co-finite. Importantly, $f$ may fail to be one-to-one, which appears to be something that you are assuming.


An important thing to remember about $\mathbb R$ with the usual topology is that it is Hausdorff. This is the central important fact for this problem. In fact, for any infinite space $X$ with the co-finite topology and any Hausdorff space $Y$, the only continuous functions $f : X \to Y$ are constant.

Proof. Suppose that $f : X \to Y$ is a nonconstant continuous function. Then there are distinct $u,v \in Y$ in the range of $f$. By Hausdorffness there are disjoint open sets $U,V \subseteq Y$ with $u \in U$ and $v \in V$. By continuity both $f^{-1}(U)$ and $f^{-1}(V)$ are open and nonempty, so they are co-finite. But then the intersection $f^{-1} (U) \cap f^{-1} (V)$ must be nonempty, which is impossible as $$f^{-1} (U) \cap f^{-1} (V) = f^{-1} ( U \cap V ) = f^{-1} ( \emptyset ) = \emptyset.$$

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A slightly different spin on vow lacks forte's answer:

Note that $\mathbb{N}$ with the co-finite topology is connected. Therefore its image under a continuous function is connected. The (non-empty) connected subsets of $\mathbb{R}$ are singletons or intervals. As the latter are uncountable, every continuous $f \colon \mathbb{N} \to \mathbb{R}$ must be constant.

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    $\begingroup$ The proof in vow lacks forte's answer essentially shows connectedness of $\mathbb{N}$ with the cofinite topology. Using the Hausdorff property is certainly more elementary than the classification of connected subsets of $\mathbb{R}$, but nevertheless I like this way of putting it. $\endgroup$ – spin Aug 28 '16 at 0:44

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