I'm presented with the following question, which I think is meant to be a precursor to material on completeness.

Let $\alpha$ be an irrational number. Show that the function $f:\mathbb{Q} \to \mathbb{Q}$ defined by $$f(x) = \begin{cases} x, & x < \alpha \\ x+1, & x > \alpha \end{cases}$$ is continuous.

Now intuitively, I feel like for $x_0<\alpha$, clearly $\lim_{x\to x_0^-}f(x) = f(\lim_{x\to x_0^-}x)$, and for $x_0>\alpha$,$\lim_{x\to x_0^+}f(x) = f(\lim_{x\to x_0^+}x)$.

The harder part to get my head around is the limits from the other direction. So for example, suppose $x_0<\alpha$, but VERY CLOSE to $\alpha$. Even though it is very close to $\alpha$, say, so that $|x_0 - \alpha| = \epsilon$, by the Archimedean Property of Natural Numbers ($\forall \epsilon > 0$ $\exists n \in \mathbb{N}$ such that $\frac{1}{n}< \epsilon$) we can find a new rational number $x$ between $\alpha$ and $x_0$ and can thus approach $x_0$ from the right, therefore confirming that $\lim_{x\to x_0^+}f(x) = f(\lim_{x\to x_0^+}x)$, proving that $f$ is indeed continuous (after using an analagous argument for $x_0>\alpha$) since $f$ is both left- and right-continuous.

Is this argument correct though? A graph of the function troubles my intuition... Can anyone make this clearer for me to get my head around?!

  • $\begingroup$ Your argument is correct. You used convergent sequences to prove continuity and in this case that can indeed be applied. $\endgroup$ – drhab Aug 19 '15 at 8:43

Alternative route to prove the continuity of $f$.

Note that the maps $g,h$ prescribed respectively by $x\mapsto x$ and $x\mapsto x+1$ are evidently continuous.

Since $\alpha\notin\mathbb Q$ we have $$\mathbb Q=[\mathbb Q\cap(-\infty,\alpha)]\cup[\mathbb Q\cap(\alpha,\infty)]$$ So for $S\subseteq\mathbb Q$ we find: $$f^{-1}(S)=[f^{-1}(S)\cap(-\infty,\alpha)]\cup[f^{-1}(S)\cap(\alpha,\infty)]=[g^{-1}(S)\cap(-\infty,\alpha)]\cup[h^{-1}(S)\cap(\alpha,\infty)]$$

If $S$ is open then the RHS denotes an open set.


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