How can you prove that the only continuous function from real number with the Euclidean topology to natural number with the cofinite topology is the constant function (without metrics)?

I tried to use disconnections and the fact that the preimage of a closed set is close if the function is finite, but I can't find anything, as N is connected with the cofinite topology...

Thank you!


For $f:\mathbb{R} \rightarrow \mathbb{N}$, the set $\{ f^{-1}(x) | x \in \mathbb{N} \}$ partitions $\mathbb{R}$ into a countable nummber of closed sets, which by a theorem of Sierpinski(?) requires the partition to have just one part.


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