Theorem 3.3 (d) Rudin

Question

This from Rudin's Principle's of Mathematical Analysis

• I'm having trouble getting an intuitive picture of this proof. Can you please show me how Rudin gets to the inequality $|s_n -s|< \frac{1}{2} |s|^2 \epsilon$ algebraically? And also a description (I don't expect you draw it out, it's fine) of what this would look like in a pictorial?? I'm guessing it has something to do with the two triangles in the third one. Also point out if I interpreted the picture wrong in any way. Thanks in advance! This sites a life saver.

Answer 1

This is a typical slick Rudin proof. The crucial algebra is, of course, putting things over a common denominator. $$\left|\frac 1{s_n} - \frac 1s\right| = \left|\frac{s-s_n}{s_ns}\right| = \frac{|s_n-s|}{|s_n||s|}.\quad (\star)$$ We know that $s_n\to s$, but how small must we make $|s_n-s|$ in order to make ($\star$) smaller than $\epsilon$? (This is the usual game in analysis.) We want to keep $|s_n|$ from getting too small (which would make $1/|s_n|$ very large), so Rudin first requires $|s_n-s|<|s|/2$ (we have $m\in\Bbb N$ so that this holds for $n\ge m$) in order to guarantee that $|s_n|>|s|/2$. Thus, whenever $n\ge m$, we'll have $$\left|\frac 1{s_n} - \frac 1s\right|<\frac{|s_n-s|}{|s|^2/2} = \frac 2{s^2}|s_n-s|.$$

Now, given any $\eta>0$, there is $N\in\Bbb N$ (depending on $\eta$, of course) so that $|s_n-s|<\eta$ whenever $n\ge N$. This will make $$\left|\frac 1{s_n} - \frac 1s\right|<\frac2{s^2}\eta.$$ Our ultimate goal is to have ($\star$) less than $\epsilon$, so how should we choose $\eta$ to guarantee $\frac2{s^2}\eta\le\epsilon$. One obvious choice, then, is to use $\eta = \frac{s^2}2\epsilon$. Finally, choosing $n\ge\max(m,N)$ will meet our desired goal.

By the way, I applaud your drawing pictures. Keep it up. I don't think it helps too much on this particular proof, but keep it up!

Answer 2

If you were trying to work this out yourself, you could bring the fractions together to get

$$ \left\lvert \frac{1}{s_n} - \frac{1}{s} \right\rvert = \left\lvert \frac{s - s_n}{s_ns} \right\rvert = \frac{1}{|s_n s|} |s_n - s|. $$

Now we know that $|s_n - s| < \varepsilon$ if $n$ is sufficiently large because that's what it means for $s_n \to s$. So what would be helpful is if we could bound $|s_n s|^{-1}$ by some constant $M$ to get

$$ \left\lvert \frac{1}{s_n} - \frac{1}{s} \right\rvert = \frac{1}{|s_n s|} |s_n - s| \le M |s_n - s| $$

and then use the usual trick of letting $|s_n - s| < \varepsilon M^{-1}$ to get $\varepsilon$ to appear on the right hand side.

Next we observe that

$$ \frac{1}{|s_n s|} \le M \iff |s_n| \ge \frac{1}{|s|M}. \tag{$*$} $$

That is, we need to bound $|s_n|$ away from $0$ by an appropriate amount. If we bound $|s_n| \ge C > 0$ and $s_n \to s$ then $|s| \ge C$ which tells us that $0 < C \le |s|$. On the other hand, we can't take $C = |s|$ because for example $s_n = 1 - 2^{-n} \to s = 1$ but $|s_n| < |s|$ for all $n$. Therefore $0 < C < |s|$.

Now it is reasonable to take $C$ right in the middle of $0$ and $|s|$, namely $C = \frac12 |s|$. We notice that for sufficiently large $n$ we do indeed have $|s_n| \ge \frac12 |s|$ since if $|s_n - s| \le \frac12 |s|$ then $|s_n| \ge \frac12 |s|$. The picture here is that if you draw a circle of radius $\frac12 |s|$ around $s$ then the closest the circle gets to $0$ is on the line from $0$ to $s$. Specifically, it intersects halfway at $\frac12 s$. Therefore if $s_n$ is in that circle then $|s_n| \ge \frac12 |s|$.

Now we can finally solve for $M$ in $(*)$:

$$ \frac12 |s| = \frac{1}{|s|M} \implies M = \frac{2}{|s|^2}. $$

And then finally take $n$ large enough so that

$$ |s_n - s| < \varepsilon/M = \frac{1}{2}|s|^2\varepsilon. $$