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Prove that $-\frac{1}{m} < a -b < \frac{1}{m}$ implies $-\frac{1}{m} < \sigma a -\sigma b < \frac{1}{m}$ for every positive integer $m$. Conclude that $\sigma \in $ Aut$(\mathbb{R}/\mathbb{Q})$ is a continuous map of $\mathbb{R}$.

proof: Let $\sigma \in$ Aut$(\mathbb{R}/\mathbb{Q}) . Then $Aut$(\mathbb{R}/\mathbb{Q})$ is the collection of automorphisms of $\mathbb{R}$ which fix $\mathbb{Q}$.

Suppose $-\frac{1}{m} < a -b < \frac{1}{m}$. Then multiply both sides by $\sigma$

So $\sigma(-\frac{1}{m}) < \sigma(a -b) < \sigma \frac{1}{m}$.

Then since $\sigma$ fixes $\mathbb{Q}$.

Then $-\frac{1}{m} < \sigma a -\sigma b < \frac{1}{m}$.

So $|\sigma a -\sigma b | < \frac{1}{m}$.

Then suppose for every $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |a - b| < \delta$

Can someone please verify I am on the right track? And I am kind of stuck on showing $\sigma$ is continuous. Thank you!

user26857
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Mahidevran
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2 Answers2

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To prove that a function $f$ is continuous you need to show that $\forall \ n\in \mathbb{N}$, whenever $|a-b|<\frac1n$ then $|f(a)-f(b)|<\frac1n$ . That's what you showed for $\sigma$.

seeker
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  • So for every $\epsilon > 0$ , there is a $\delta = \frac{1}{m}$ such that whenever $|a- b | < \frac{1}{m}$ then $|\sigma(a) - \sigma(b)| < \epsilon$.? – Mahidevran Dec 10 '15 at 04:13
  • @Mahidevran yes. when you say that $\forall \epsilon>0\ \exists \delta>0$, then since $\delta>0$ hence by the Archimedean Property $\exists n\in \mathbb{N}$ such that $\frac 1n <\delta$. – seeker Dec 10 '15 at 04:14
  • Dan's right that you phrased this poorly. You don't need to show the property in the answer to have continuity -- plenty of continuous functions don't satisfy it ($x \mapsto 2x$ for example) -- but rather it implies continuity. It's a sufficient condition, but not necessary. – epimorphic Dec 28 '15 at 01:28
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    He didn't show continuity, because he didn't show that $\sigma$ preserves the ordering. Proving that is the main step of this problem. – Elle Najt Dec 28 '15 at 04:06
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You first need to prove that $\sigma$ preserves the ordering.

This follows from the observation that $x$ so that $x \geq 0$ are exactly the $x$ which are squares in $\mathbb{R}$.

Elle Najt
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