User talk:Ringell

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Cheers! prime mover (talk) 12:42, 21 August 2019 (EDT)

Question about Metric Space Continuity by Open Ball

You added this comment:

(*) Only consider this case, since otherwise $\map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ is vacuously true.

I don't understand. What the proof is purporting to do is demonstrate that, given $\map {d_1} {x, a} < \delta$, it can be proved that $\map {d_2} {\map f x, \map f a} < \epsilon$.

That is accomplished by the 3 lines given; that being done, the proof is complete.

What is the "otherwise"?

Also note that the extra lines of clarification at the start of the proof do nothing more than merely restate exactly what has been stated at the top of each section, so I need to be convinced that we need to keep them. --prime mover (talk) 08:11, 30 November 2020 (UTC)

My intent with these things, was to make the implicit assumptions (as I now realise come from a personal not universal understanding or making sense of proofs) in the proof a bit clearer. But I agree that it is too much. It adds unnecessary clutter, especially if done on a whole site level. It also forces a certain understanding of implications and of proofing in general, which is unhelpful and confusing. Sorry for the extra work and thank you for taking the extra time to contact me.

Regarding

(*) Only consider this case, since otherwise $\map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ is vacuously true.

the idea was that, if $\map {d_1} {x, a} < \delta$ were false, the implication is always true. Which is unhelpful, since this is the case for every implication. I think I was just overly aware of this fact, since I previously saw this a, b case differentiation in the ϵ-δ Definition implies Definition by Limits on the Metric Space Continuity by Epsilon-Delta page, where the b case does not need to be proven since it is vacuously true and especially adds an unnecessary dependency to one of the metric space axioms. Without this dependency the proof would also hold for pseudo metric spaces and not just metric spaces.

The extra lines of clarification at the start were intended to make clear, where the $\delta$ comes from, since I was initially shortly confused and thought I would make this clearer for the next readers convenience. But I think, especially on a whole site level, this is unecessary clutter, since it is something more experienced proof readers are probably well aware of.

Kind regards ----Ringell (talk) 07:48, 1 December 2020 (UTC)

Yeah, good call. Okay, check the amendment that I just made -- that should make it clearer exactly where the $\delta$ comes from. --prime mover (talk) 09:17, 1 December 2020 (UTC)

Thank you, the rework reads very nicely. --Ringell (talk) 18:52, 1 December 2020 (UTC)