Let $\alpha\in\mathbb R$ be an irrational number. Show that the function $f : \mathbb Q \to\mathbb Q$ is continuous, where $f$ is given by $f(x) = x$ for $x < \alpha$ and $x + 1$ for $x > \alpha$.
I'm not sure how to go about this, I've been trying to use the fact that every rational number has a sequence of irrationals converging to it, but it doesn't seem to go anywhere. Any help would be appreciated. Thanks!
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