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I was asked to show that the limit of $z^2 = -1$ as $z$ approaches $I$ and my working out is shown below. I am stuck on a few things.

Question 1: Why do we choose $\delta$ to be the infimum? Based on my understanding, since $\epsilon$ is arbitarily small, the infimum will always be $\frac{\epsilon}{3} < 1$ and to me, choosing an infimum is pointless.

Question 2: Why do we substitute 2 different $\delta$ values in the last step. Namely we substitute $\delta = 1$ and $\delta = \frac{\epsilon}{3}$. Shouldn't we pick the one that is an infimum and stick to it?

Question 3: The notation for the infimum confuses me. Some textbooks show it using {} brackets as to indicate a set and others along with my lecturer use () to indicate an interval. Which is it supposed to be in the context of $\epsilon-\delta$ proofs?

Let $\epsilon$ > $0$.

By definition $|z - i| < \delta$.

Now: $|z^2+1|= (z - i)(z + i) = (z - i)(z - i + 2i)$

It is sufficient to choose $\delta = 1$; i.e., $|z - i| < 1$

By the triangle inequality:

$|z - i||z - i + 2i| < |z - i|(|z - i|+2)$

Since $|z - i| < \delta$ implies $|z^2+1|< \delta(\delta +2)$

We choose $\delta = \inf(\frac{\epsilon}{3}, 1)$.

Then: $|z-i|(|z-i|+2) < \frac{\epsilon}{3} (1+2) = \epsilon$.

Beerus
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Oofy2000
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    Using $\inf$ there is very weird. For any set that has a minimum value the infimum coincides with the minimum. That is true, in particular for finite sets. So $\min{\frac{\epsilon}{3},1}$ would be more standard, even though of course the other notation means the same. – Gold May 12 '24 at 14:07
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    Not marking as a duplicate, but https://math.stackexchange.com/questions/966789/why-do-we-need-min-to-choose-delta may be of interest. – Andrew D. Hwang May 12 '24 at 15:59
  • @AndrewD.Hwang thank you for sharing. It makes a lot more sense now. The one thing I would like to confirm is if saying “it is sufficient to choose $\delta$ = min (a, b)” is equivalent to saying $\delta$ < a, b” since a and b are the largest possible values and choosing them to be the minimum is sufficient for our proof. – Oofy2000 May 12 '24 at 23:00
  • (i) The two are not equivalent: Writing "$\delta \leq a$, $b$" would mean, to me, $\delta \leq \min{a, b}$. But (ii) In this type of proof, taking $\delta $ "smaller than necessary" is all right, so in the spirit of your question, the lack of logical equivalence does not create a logical gap. – Andrew D. Hwang May 13 '24 at 00:32
  • @AndrewD.Hwang So in the case of an epsilon-delta proof (using the case above as an example), shouldn't we write |−|(|−|+2)< (+2) <= $\frac{\epsilon}{3}(1+2)$ <= $\epsilon$ and NOT |−|(|−|+2)< (+2) < $\frac{\epsilon}{3}(1+2)$ < $\epsilon$ – Oofy2000 May 13 '24 at 01:20
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    Without checking too carefully, probably yes. If it helps, there is still a strict inequality before $\delta(\delta+2)$, so the whole daisy-chain of inequalities is strict and the definition has been checked. – Andrew D. Hwang May 13 '24 at 11:45

1 Answers1

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I assume the problem statement is supposed to be: Show that $z^2$ approaches $-1$ as $z$ approaches $i$?

The proof then roughly has the following form: Given any $\epsilon > 0$, we choose $\delta := \min\{1,\frac{\epsilon}{3}\}$ and show that for any $z$ with $|z-(-1)| < \delta$ we get $|z^2-i| < \epsilon$.

With this in mind, to your questions:

  1. In my opinion, this is mostly a matter of taste. But if you want to be completely formal (and only apply the standard $\epsilon$-$\delta$ definition for convergence), choosing $\delta := \min\{1,\frac{\epsilon}{3}\}$ is the right way of doing it, since you have to find a $\delta$ for any $\epsilon$. Of course, it is not to hard to see that, in fact, it suffices to only consider "sufficiently small" $\epsilon$. Once, one understands this, one can alternatively start the proof by saying: "Given any $\epsilon > 0$ (without loss of generality $\epsilon < 1$), we choose $\delta := \frac{\epsilon}{3}$..."

  2. Since $\delta$ is defined as the minimum of both values, you can use both estimates $\delta < 1$ and $\delta < \frac{\epsilon}{3}$ wherever you want. And using each one once here, happens to give the nice upper bound $< \epsilon$.

  3. The notation $\inf(\frac{\epsilon}{3},1)$ seems very weird to me. I would always use $\inf\{\frac{\epsilon}{3},1\}$ (or, even better, $\min\{\frac{\epsilon}{3},1\}$) here. If $(\frac{\epsilon}{3},1)$ refers to the interval between $\frac{\epsilon}{3}$ and $1$ in the first notation (which is the standard interpretation), this isn't even correct in the case that $\frac{\epsilon}{3}>1$ (and certainly confusing in any case).

GraffL
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  • Thank you for your reply. That clears up a lot. One thing though regarding $\epsilon$ < 1 for instance. Are we able to say this since we know the largest possible value it can take on is 1? Also, if we chose $\delta$ < $\frac {\epsilon}{3}$ doesn’t that mean $\delta$ is not <1? – Oofy2000 May 12 '24 at 14:03
  • Do you mean, why we can assume $\epsilon < 1$ without loss of generality? Intuitively, this is because $f(x)$ converging to $f(y)$ for $x$ converging to $y$ just means: $f(x)$ gets arbitrarily close to $f(y)$ if $x$ gets close to $y$. Hence, we are only interested in $x$ which are close to $y$. Though, one has to first formally prove such a lemma, before one may use it (proving such a statement is a nice exercise in applying the $\epsilon$-$\delta$-definition in a more abstract setting. But maybe a bit much to start with). When in doubt, you can always just work with general $\epsilon$ – GraffL May 12 '24 at 14:17
  • Sorry, I meant in regards to your point 2. By choosing $\delta$ = min{$\frac{\epsilon}{3}$, 1}, are we not choosing one or the other and not both? – Oofy2000 May 12 '24 at 14:21
  • Also, by setting the minimum value, does that mean the largest value our delta can take is 1, and so $\delta$ < 1? – Oofy2000 May 12 '24 at 14:22
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    We are choosing the minimum. Whether this means $\delta = \frac{\epsilon}{3}$ or $\delta = 1$ depends on the value of $\epsilon$. The point is: It doesn't matter for the rest of the proof which of the cases holds. We only use $\delta \leq \frac{\epsilon}{3}$ and $\delta \leq 1$ which is true in any case. – GraffL May 12 '24 at 14:26
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    Alternatively, you could also do a case distinction, i.e. split your proof in two cases: "If $\epsilon \geq 3$ we choose $\delta = 1$ and .... Otherwise, we choose $\delta = \frac{\epsilon}{3}$ and ...." Then, in each of the two cases, you know exactly what $\delta$ is (and can use this). This would yield an equally valid proof, just a bit longer. – GraffL May 12 '24 at 14:29
  • Let us continue this discussion in chat. – Oofy2000 May 13 '24 at 00:36