# User talk:JRSpriggs

Resources:

Archives:

## Axiom of regularity and consistency of ZF without it

Hello, why did you revert my edit that clarified how the question of consistency of ZF without regularity is relevant to proving that there is no set of all sets? --Alexey Muranov (talk) 09:01, 18 July 2018 (UTC)

Currently the phrase "However, if the ZF axioms without regularity were already inconsistent, then adding regularity would not make them consistent." is out of place, because consistency of ZF was not a subject of the paragraph, and in the whole section only inconsistency of Naive set theory was mentioned. Moreover it is in any case obvious that adding axioms to an inconsistent theory will not make it consistent. So, now we have an obvious remark without any context that could justify it. I will restore my edit that is in line with the first part of the paragraph (discussion of a proof that there is no set of all sets). Removing the phrase altogether would also be fine IMO, but you opposed it previously. Otherwise the whole paragraph needs to be rewritten to make the obvious remarks on the question of consistency of ZF somehow relevant. --Alexey Muranov (talk) 09:21, 18 July 2018 (UTC)

My original edit summary explained why I put the sentence back in. It said "restore important (even if obvious to some) sentence". The fact that it was obvious to you and to me does not imply that it will be obvious to all readers. I expect that some readers of this section will be thinking something like << This section is about how to eliminate the Russell paradox. Apparently, we can just say that it is not allowed in our set theory. The axiom of regularity is how we say that. >>. Although I was not the author of that sentence, I appreciated that it was there to counter-act that mistaken thinking. That is why I put it back. Your attempt to change what I did (thinking that you know better what I want than I do) does not disabuse the reader of this error. JRSpriggs (talk) 17:17, 19 July 2018 (UTC)
I am sorry, but I think that what you expect about what the reader will think is not a good enough justification to add a remark that is irrelevant to the context (and thus is likely to confuse readers who will not think what you expect they will think). It confused me a bit, and I started to look how it could be relevant to what was said before. Russell's paradox appears in the Naive set theory, not in ZF, as is clearly stated in this section. It is even explained what remains of it in ZF (a theorem on the absence of the "universal set"). If ZF without regularity is inconsistent, then everything is a theorem, as I explained in my edit, and there is not much do discuss. --Alexey Muranov (talk) 22:25, 19 July 2018 (UTC)
It is now in a separate paragraph and explained more clearly, so please leave it alone. JRSpriggs (talk) 03:14, 20 July 2018 (UTC)
Dear JRSpriggs, to me the added paragraph still looks out of place: the possibility of inconsistency of ZF was not a topic of this section until that point and it seems to be irrelevant to the rest. It is already stated that (literal) Russell's paradox does not happen in ZF, so even if ZF is inconsistent, it is not "because" of Russell's paradox, end of story. It also seems that calling consequences "undesirable" is mathematically meaningless, unless there is some context that explains what is "desired." (Models with "undesirable properties" in the first paragraph make more sense to me: excluding "pathological" models means adding consequences.)
You wrote: "if ZF without regularity is extended by adding regularity to get ZF, then any contradiction (such as Russell's paradox) which followed from the original theory [...]". The mention of Russell's paradox in parentheses is rather confusing if not wrong: there is no literal Russell's paradox in ZF.
To make that added remark somehow on topic, I can only reword it thus: "There is no Russell's paradox in ZF with or without regularity, but if it were there without regularity, we would not be able to avoid it by adding regularity back." To me this qualifies as a moderately funny mathematical joke.
In general, this added remark seems to be on the way to making a list of all possible silly ideas a reader can come up with and warning against them all. As I keep saying, the question of consistency of ZF was not even mentioned before, so a hypothetical reader who, instead of reading what is written, starts making strange hypotheses about fixing hypothetical inconsistencies in ZF can be honestly left on their own, IMO.
I suggest removing the sentence which was the original cause of dispute and the added paragraph, and instead I'll add a paragraph along the lines of what you suggest, but about the Naive set theory instead of ZF, because this will be at least on topic (Russell's paradox), though somewhat redundant IMO. --Alexey Muranov (talk) 10:50, 20 July 2018 (UTC)
I edited it, and then edited again, because I realised that trying to follow your line of thought I got confused myself again, and that it is better to be clear that this remark is an aside.
Even if we assume that someone will imagine the possibility of avoiding Russell's paradox by forbidding the set of all sets, that person would not need to add axioms for this: the set of all sets is already forbidden by Rassell's paradox!
IMO you are making several unjustified assumptions: (1) that the reader is likely to be confused in that particular way, (2) that you suggested explanation will likely address the root cause of their confusion (otherwise it would be better to avoid giving any explanations and to let the reader think on their own), (3) that that reader will not be confused about some more important points (otherwise why to turn their attention, and also the attention of those who were not initially confused, to that particular point?). --Alexey Muranov (talk) 13:36, 20 July 2018 (UTC)
1. We do not know that Russell's paradox does not occur in ZF, merely that the easy derivation of it in naive set theory fails in ZF.
2. I think that we can safely say that a contradiction would be an undesireable consequence of any theory.
Again, please leave my paragraph alone! It is meaningful and true and relevant. JRSpriggs (talk) 04:49, 21 July 2018 (UTC)
What do you mean by Russell's paradox? I meant a certain derivation of contradiction in Naive set theory, which does not work in ZF because axioms are not the same. However, it can be "turned around" (as mentioned in the section we are discussing), in which case it becomes a proof that ZF excludes the existence of "set of all sets" -- as well as Naive set theory does. (This by itself cannot be called a "paradox" IMO.) --Alexey Muranov (talk) 11:52, 21 July 2018 (UTC)
If someone found a way to prove in ZF without regularity that the class of all sets was a set, then one could use the axiom of separation to show that the "Russell set" existed and get the paradox that way. This seems very unlikely since if it were possible, it probably would already have been done. But one cannot prove (in ZF) that it is impossible, unless it is possible. JRSpriggs (talk) 04:08, 22 July 2018 (UTC)
Sorry, i still fail to see which hypothetical derivations of contradiction you call "Russell's paradox." If someone derived a contradiction in ZF and used it to prove the existence of Russell's set, from which a contradiction can be derived, could that be called Russell's paradox in ZF, in your opinion? I hope not, otherwise it would seem like you can call "Russell's paradox" any derivation of contradiction in ZF. --Alexey Muranov (talk) 08:08, 22 July 2018 (UTC)
I clarified in the preceding paragraph that ZF without the axiom of regularity already prohibits the set of all sets, so your added paragraph seems to be clearly redundant now, please take a look. I still do not see how it is relevant: your explanation seemed to be that you expect a substantial part of readers to be mistaken in some particular way, and you suppose that you paragraph would be helpful, but I do not see what makes you think so. I am also fuzzy about the way you use the term "Russell's paradox" in relation to ZF. --Alexey Muranov (talk) 08:29, 22 July 2018 (UTC)

───────────────────────── You said "... Russell's paradox yields a proof that there is no "set of all sets" using the axiom schema of separation ...". If this is how you show that Russell's paradox cannot occur, then you are begging the question. JRSpriggs (talk) 07:09, 23 July 2018 (UTC)

Excuse me, but you seem to have ignored what i was asking in my last comments. Regarding yours "If this is how you show that Russell's paradox cannot occur, [...]", it seems to be "vacuously true," because its premise is apparently false, though I still do not see what exactly you mean by Russell's paradox, and I hesitate to comment on the truth value of sentences that i do not quite understand (like "this sentence is false")...
Could you please (1) clarify which exactly hypothetical derivations of contradiction in ZF you would call Russell's paradox (see my question with more details above), (2) confirm whether you agree that your added paragraph is now clearly redundant and somewhat off topic. --Alexey Muranov (talk) 18:27, 23 July 2018 (UTC)
I already said "If someone found a way to prove in ZF without regularity that the class of all sets was a set, then one could use the axiom of separation to show that the "Russell set" existed and get the paradox that way.". I hesitate to give a more precise answer because I think that classifying derivations of a contradiction as one kind or another is a fool's errand. That is why I put the reference to Russell's paradox in parentheses. I think that given a derivation of a contradiction, one could either manipulate it to make it a instance of Russell's paradox or manipulate it to make it not an instance. Here, by an instance, I mean that the first obviously absurd step (other than possibly the set of all sets) is the set of all sets not containing themselves.
No, I do not agree. JRSpriggs (talk) 04:57, 24 July 2018 (UTC)
(1) "[...] I think that classifying derivations of a contradiction as one kind or another is a fool's errand." -- isn't this what you need to accomplish to be understood when referring to "Russell's paradox" in ZF, especially since the literal Russell's paradox does not occur? So far your use of "Russell's paradox" in parentheses looks confusing to me. I would rather take care of this, than of a hypothetical confusion of a reader about something that was never claimed or implied.
(2) Could you explain how it is relevant, given that ZF without regularity already prohibits the set of all set? (How is it relevant, an example of how one cannot construct a consistent theory if ZF without regularity is inconsistent? There are a lot of ways how one cannot do whatever you like.) --Alexey Muranov (talk) 20:17, 24 July 2018 (UTC)
It seems to me that you are conflating ${\displaystyle \vdash \lnot ((R\in R)\iff \lnot (R\in R))}$ with ${\displaystyle \lnot \vdash ((R\in R)\iff \lnot (R\in R))}$. JRSpriggs (talk) 21:30, 24 July 2018 (UTC)
I do not.
You do not do much (besides your very first reply) to explain how your added paragraph is relevant, and sometimes you avoid answering direct questions (my question in (2) just above, for example). Don't you agree that if someone claims that something is relevant and has to be included, the burden of justification should ultimately be on him? (I hardly see how it could be possible in principle to justify that something is irrelevant, if my efforts above do not count.) --Alexey Muranov (talk) 14:57, 25 July 2018 (UTC)

───────────────────────── The first occurrence of the section on Russell's paradox in axiom of regularity was at [1]. It was clearly motivated by showing that the axiom of regularity cannot rescue ZF from contradictions, contrary to a common myth. You say "ZF without the axiom of regularity already prohibits such a universal set". But what do you mean by "prohibits"? Do you mean that one can prove that the universal set does not exist or do you mean that one cannot prove that it does exist? All that you justify is the former. My paragraph is clarifying that regularity does not help us with the latter, a point which you do not make in your paragraph. JRSpriggs (talk) 21:35, 26 July 2018 (UTC)

Thanks for the link to the original version, it makes much more sense IMO than the paragraph we are discussing. The current paragraph is included without any context and thus seemingly off-topic, but apparently it tries to address a "common misconception" (which was the motivation for the origin of this section and which was clearly stated in the initial version) without providing any evidence that such a misconception is common (I asked you about some such justification above) and without even mentioning it. If such a misconception exists, it is indeed interesting to mention it in the article providing some references (like what the original version did, mentioning Everything and More (book)), because it is the existence of the misconception that is worth discussion, not so much some trivial proof that it is a misconception. In the current version of the section, at least after my edit, it is clear that adding an axiom to be able to prove that there is no set of all sets is already useless (as it can already be proved), no need to look further.
IMO a solution would be to restore a big part of the original version of the section as a paragraph in the current section. I would even vote to revert the section to the initial version completely, as besides the discussion of the "common misconception" there is basically no relation between Regularity and Russell's paradox.
Note: I wrote the above under the assumption that the mentioned misconception is indeed common, which i know nothing about. If it is not, then it should be called "curious misconception," or "Regularity and Russell's paradox in popular culture," or something else, and in such case no explanation would be needed i think.
To be consistent, I used the word "prohibits" in the same sense in which it was used in the previous sentence (by someone else). --Alexey Muranov (talk) 07:57, 28 July 2018 (UTC)

## Cardinality confusion

In regards to your revert edit in the cardinality article, their are two interpretations of one to one, one (each member of set) to one (for each or some members of set) or one to one for all of each set once only for each member. Additionally bijection is claimed by different authors as being determined by one to one, or by either one of two versions of a ≤ test and a ≥ test thus 'bi…'

Victor Kosko (talk) 00:07, 27 July 2018 (UTC)


## Equivalents to Axiom of Choice

Whoops! You are absolutely right, thanks for catching that; it was meant to say "there is either an injection or a surjection from any nonempty A to B" but the equivalence of that to the first sentence is trivial and probably not worth adding separately. 2601:42:0:4C76:B8C1:CEBA:5ACF:59A2 (talk) 08:55, 9 September 2018 (UTC)

## Multilinearity

Hello! I noticed you made an edit to the article on tensors (which has since been re-worked by Purgy Purgatorio), where you removed "multilinear" because, as you said, all tensors are "multilinear": https://en.wikipedia.org/w/index.php?title=Tensor&type=revision&diff=860649481&oldid=860618369&diffmode=source

It seems to me that "multilinear" is an ambiguous term, due to the "multi" part of the word. I could be wrong about this, but it seems that, usually, "multi" implies "more than one", and sometimes it implies "an arbitrary whole number". In an article such as "Multilinear map", this ambiguity leads to apparent contradictions in the article.

If all tensors are unambiguously multilinear, then (geometric) scalars should be multilinear, and it would seem that "Multilinear map" should be reworked to clearly explain this general definition and include the trivial example of (geometric) scalars. If this "multilinear" terminology is actually ambiguous, then it seems that that should be explained both in the "Multilinear map" article and the "tensors" article.

Zeroparallax (talk) 05:45, 25 September 2018 (UTC)