1

On pages 327 and 328 in R. Courant's book What Is Mathematics the global continuity of the function $x \to \frac{1}{1+x^{2}}$ is shown. But I don't understand that proof which runs essentially as follows:

On finding $$|f(x) - f(c)| = |x-c|\frac{|x+c|}{(1+x^{2})(1+c^{2})}$$ he says we can restrict $x$ to a fixed interval $|x| \leq M$ with $M$ an arbitrarily selected number. Then he says for $|x|, |c| \leq M$ we have $$|f(x) - f(c)| \leq |x-c|2M < \delta 2M\leq \epsilon$$.

Two things that I don't understand here are:

1) Why can we first choose $M$?

2) Why can we ignore the denominator?

What have I missed?

299792458
  • 507
  • 7
  • 18
Yes
  • 20,719
  • 1
    The arbitrary choice of $M$ is necessary for the general proof. You ignore the denominator to get the crude (yet still sufficient) upper bound. – A.E Sep 02 '14 at 05:59

1 Answers1

1

$|x+c| \leq |x| + |c| \leq M + M = 2M$, and $1+x^2 > 1$, $1 + c^2 >1$, and take $\delta = \dfrac{\epsilon}{2M}$, then conclusion follows.

DeepSea
  • 77,651
  • I am not sure if so then $M$ is independent or dependent? Will you please show me the logical form with quantifiers appearing? – Yes Sep 02 '14 at 06:06
  • @Comeseeconquer Fix $M$ (for all $M$ - we want $M$ large). Let an opponent choose $\epsilon$ (for all $\epsilon \gt 0$ - our opponent wants $\epsilon$ small) then we can choose $\delta$ (there exists $\delta$) - note that big $M$ and small $\epsilon$ mean that $\delta$ is small - the bigger the challenge, the smaller $\delta$ has to be. – Mark Bennet Sep 02 '14 at 06:14
  • @MarkBennet: Thanks. So it begins with: For every real $M > 0$ and every $\epsilon >0$ and every real $c$ there is a $\delta > 0$ ...? That is, $M$ is independent in this context? – Yes Sep 02 '14 at 06:16
  • @Comeseeconquer Indeed, and that is important, because it means we can establish continuity over as large a region as we like. – Mark Bennet Sep 02 '14 at 06:18
  • @MarkBennet: Thank you so much, I think I get it. One more question: So can I prove the global continuity of $1/x$ by the following: Fix any real $M > 0.$ Then for $|x|, |c| \geq M$ we have $$\frac{|x-c|}{|xc|} \leq \frac{|x-c|}{M^{2}} < \frac{\delta}{M^{2}} \leq \epsilon,$$ so $\delta := M^{2}\epsilon$ suffices? – Yes Sep 02 '14 at 06:21