= Solution
Every nonempty <open set> in the given <topology> contains $1$. Suppose a <continuous map> $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset\mathbb R$ containing these respective values. By <continuity>, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty <open sets> in $\mathbb N$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus \b[every such continuous map is constant]. Conversely, constant maps are <continuous>. This is the <continuous real maps from the initial-segment topology> argument, using the <Hausdorff> separation of distinct real points.
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