If image of such continuous function is infinite then I can't actually construct such map..and also I can't exactly claim that this map is constant..
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1The image has to be compact, since $\left[ 0, 1 \right]$ is a compact interval. So, now the question it what are all the compact sets in $\mathbb{N}$ with the cofinite topology? – Aniruddha Deshmukh Oct 10 '21 at 04:26
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What kind of characterization of the functions are you looking for? I would say that, since a function is continuous iff the preimage of any cofinite set is open, then it's all the functions such that the preimage of any point is closed. – Joe Oct 10 '21 at 05:00
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1See also: Proving function between topological spaces must be constant and Prove that $\mathbb{N}$ with cofinite topology is not path-connected space. – Martin Sleziak Oct 10 '21 at 06:31
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1A theorem by Sierpiński says that a compact connected Hausdorff space (aka a continuum) cannot be a union of countably many pairwise disjoint and non-empty closed sets (unless there is only one set $X$ in that family). See also here. – Henno Brandsma Oct 10 '21 at 07:02
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Let $f$ be continuous, then $f^{-1}(a)$ is closed for every $a\in\mathbb N$. On the other hand, if this condition holds, then the inverse image of any finite subset will be a finite union of closed and hence also closed, which proves $f$ is continuous. Hence such a map is equivalent to decompose $[0, 1]$ to a disjoint union of closed subsets.
Such a decomposition has to be trivial by this question: Is $[0,1]$ a countable disjoint union of closed sets?
Therefore the only continuous maps from $[0, 1]$ to $\mathbb N$ are constants.
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