1
$\begingroup$

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?!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.