User talk:Dominic Mayers (old)

Welcome!

Hello, Dominic Mayers, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few links to pages you might find helpful:

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Please remember to sign your messages on talk pages by typing four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or , and a volunteer should respond shortly. Again, welcome! RJFJR (talk) 17:47, 6 November 2017 (UTC)

About transportability

I finally read the proof by Marshall et al. My comment on it and on transportability in general is that what is a transportable sentence depends heavily on the choice of individual constants used in the first order language that defines the possible sentences. This might be obvious, but worth keeping in mind nevertheless. In Marshall et al., there is a first set of individual constants to be interpreted as elements $Y_{i}$ of a scale set, another set of constants to be interpreted as the base sets $X_{i}$ and a constant to be interpreted as the universe of sets over the base sets $X_{i}$ of rank less than or equal to $n$. The two first sets of constants are the expected constants of a structure: the base sets and the operations (or relations). The universe of sets is only added in Marshall et al. as a flexible way to restrict the range of the variables. For example, when $n=0$, we have the cases where the variables range over the base sets. This is a reasonable choice of constants given the definition of structures in Bourbaki and the fact that the purpose of the sentences is to define species of structures.

However, the definition of transportability does not have to be tightly linked to a specific notion of structures. It can be defined for any first order language with equality that has membership as binary predicate, some constants for the base sets, other constants for other sets defined over the base sets and one or more constant for universes of sets over the base sets, which we also use to restrict the range of the variables. This is a generalization of the language ${\cal {L}}$ defined in Marshall et al. In Marshall et al., the other sets (beside the universe of sets and the base sets) are elements of some scale set, but the language can be generalized to any other way to identify sets over the base sets, which lead to different notions of structures. The universe of sets, which is used to restrict the range of the variables, can also be generalized, for example, by allowing different universe of sets and thus different kind of variables. We are not suggesting that a new notion of structures should be used in practice, but it is nevertheless useful conceptually to consider this generalization.

This is an attempt that failed. We present it, because it was the base for the discussion below with Boris Tsirelson. A different approach is provided after the discussion. In what follows, a base of sets is a set of base sets over which we define a structure. In Bourbaki, for any pair of base of sets, the canonical extensions to scale sets of any bijections between the two base of sets are expected to be themselves bijections. Similarly, a language is a transportable language or simply transportable if, for any pair of base of sets, the canonical extensions to the universe of sets and to the sets in the universe of sets of any bijections between the two base of sets are themselves bijections. Note that Bourbaki does not define such canonical extensions, but it is not hard to do that. However, these canonical extensions are not necessarily bijections in general. In fact, the canonical extension on the universe of set is not well defined. Each individual canonical extension is well defined and a bijection, but the sets might be non disjoint and the canonical extensions on two non disjoint sets might be different on their intersection. That is why this attempt fails.

The interesting point is that, given a transportable language ${\cal {L}}$, every sentence in ${\cal {L}}$ is transportable. The key idea of the proof is simply that every occurrence of $x\in y$ is equivalent to $F(x)\in F(y)$, where $F$ is the canonical extension. The converse is ill defined or trivially true if it means that every transportable sentence of ${\cal {L}}$ is a sentence in ${\cal {L}}$. This works if we have the extra constraint that $y$ is not an element of a base set. The problem, again, is that, as we explained above, the existence of the canonical extension $F$ is itself dubious.

The next statement is clearly wrong, but again we keep it because it was the basis for the discussion below. For example, if we restrict ourselves to base of sets with a single principal base set, then the language is transportable. This is interesting, given that, when we allow auxiliary base sets, the language becomes very flexible. One can even simulate many principal base sets using a mapping from the single principal base set to a set of identifiers for the simulated principal base sets.

What's going on? How comes it is so complicated in Marshall et al. and only a sub-language of ${\cal {L}}$, namely the sentences (equivalent to a sentence) of type theory, are transportable? Note that, in Marshall the proof is also relative to sentences in ${\cal {L}}$. So, if all the sentences of ${\cal {L}}$ were transportable, the converse will be trivial, just as we discussed above. Therefore, the key question is why, in their case, only the sentences of a sub-language are transportable. Look at the basic idea of the proof. Why would it fail in their case? Because the canonical extension on a set in the universe of sets is not necessarily a bijection. If it is not a bijection, then the argument of the simple proof fails.

Note that, if we consider sentences of type theory, then the simple proof does not fail, because then each variable runs over one type, all the "evil" sets are eliminated from the universe of sets, and the canonical extension is a bijection on all sets of the restricted universe. In type theory, we have a separate universe of sets for each type, but we can view the disjoint union of these universes as the single universe. Therefore, in a way, what using type theory does for us is adding a restriction on the universe of sets so that the language becomes transportable.

There are other ways to make a language transportable. One is to modify the notion of isomorphisms itself, by requiring that there is also a bijection on the union of the principal base sets. To avoid confusion, let's call them global isomorphisms. The consequence will be a weaker notion of transportability: there are less global isomorphisms and thus being invariant under global isomorphisms is a weaker statement. It could very well be that this weaker notion of transportability is less useful in most practical situations. If that is the case, then this is a strong point in support of type theory, because, essentially Marshall et al. have shown that type theory is the minimal constraint that must be imposed on the universe of sets to make the language transportable, with the current notion of isomorphism.

Dominic Mayers (talk) 22:18, 26 March 2018 (UTC)

So, what about transportability of such sentences as "$\exists x,y\in X\;\;x\in y$", "$\exists x\in P(X)\;\exists y\in X\;\;x\in y$", "$\exists x\in P(X)\;\exists y\in X\times X\;\;x\in y$" etc? Boris Tsirelson (talk) 18:00, 27 March 2018 (UTC)

As required in Marshall et al., ``the quantifiers of the language are all relativized to this constant $r$." The constant $r$ is the universe of sets, which only contains sets that are defined over the base sets. I am not sure if this answers your question, but your quantifiers aren't relativized to the universe of sets. Dominic Mayers (talk) 19:04, 27 March 2018 (UTC)

I am trying to reformulate your examples by relativizing the quantifiers to the universe of sets and then see what is going on. Dominic Mayers (talk) 19:14, 27 March 2018 (UTC)

Yes, please. As I wrote, I did not read that article in detail; you did, please translate when needed. And of course, I ask about your, weaker transportability; for the usual transportability the answers are "no", "no", "no", ... Boris Tsirelson (talk) 19:29, 27 March 2018 (UTC)

In the third example, it wouldn't be fair if the issue was that the sentence was not interpreted in a consistent manner in terms of how a pair corresponds to a set. For example, if $y=(x_{1},x_{2})$ is defined as $\{\{x_{1}\},\{x_{1},x_{2}\}\}$, then the same rule must be applied in a consistent manner. This definition is the usual definition of pairs, but it does not matter what definition is used, the set used to represent a pair will be in the universe of sets and there will be a canonical extension on it, because it would make no sense that a definition of pairs does not have a canonical extension on it. So, it is transportable and not only weakly transportable. I wanted to clarify this first. It's good that you referred to how we construct pairs, because it is implicit in the paper of Marshall et al. and also in my notes that the cartesian product is defined that way. On the other hand, it means that the cartesian product, in this context, is simply syntactic sugar to write complex examples that do not use a cartesian product. The first two examples are counter examples, but I think that I can take care of that by slightly modifying the definition of transportable language. This is good. Dominic Mayers (talk) 22:32, 27 March 2018 (UTC)

Ah, yes, if $y=(x_{1},x_{2})$ is defined as $\{\{x_{1}\},\{x_{1},x_{2}\}\}$, then $y\in P(P(X))$ and so, the relation $x\in y$ obeys the types (and therefore is transportable). "but it does not matter what definition is used" — Really? Try such definition: $\{\{\{x_{1}\},\{x_{1},x_{2}\}\}\};$ this time $(x,y)\in P(P(P(X)))$. As far as I remember, Bourbaki does not prefer one implementation of pairs to others; if so, then this transportability question is undecidable in his system. Boris Tsirelson (talk) 05:07, 28 March 2018 (UTC)

We programmers know that types may be checked statically or dynamically. The same can be made in mathematics. ZFC is like a RISC machine rather than i432. (For me i432 looked much better, but...) Static types are an implicit de facto standard in mathematics (except maybe set theory). If you like, you may design dynamic type checking in mathematics. Recall that some (probably, Bourbaki) define a function as a triple (domain,codomain,graph) rather than just graph. Similarly you may redefine P(X) as consisting of pairs (X,A) such that $\forall x\in A\;\;x\in X$ rather than of just A, treating the first element X as a type tag (like RTTI). But I'm afraid, others are not motivated to follow... Boris Tsirelson (talk) 10:52, 28 March 2018 (UTC)

Do a social experiment: ask (say) an algebraist some innocent-looking question about a field such that $0\in 1.$ I guess, he/she will ask you, is "$0\in 1$" a typo, a joke, or what? And if you'll say "you know, everything is a set; why not say 0 belongs to 1?", you'll face bewilderment or laughter. This is the de facto standard in action. Boris Tsirelson (talk) 11:53, 28 March 2018 (UTC)

Do you mean "This time $y\in P(P(P(X)))$"? Dominic Mayers (talk) 12:38, 28 March 2018 (UTC)

Oops, you are right, it should be "$(x_{1},x_{2})\in P(P(P(X)))$". Boris Tsirelson (talk) 13:06, 28 March 2018 (UTC)

Back to the social experiment... Asking an algebraist about a field such that $0\in 1$ is like asking a music fan, does he prefer sound files starting with zero bit, or bit one. Boris Tsirelson (talk) 14:19, 28 March 2018 (UTC)

In some of your arguments, you seem to equate "belongs to type theory" with "transportable". The theorem of Marshall et al. might be useful, but I would be careful, if you don't understand the paper. They have assumptions, some are explicit others are not. Therefore, just to be on the safe side, let's try to understand the issue using the basic definition of transportability, which does not depend on type theory. The usual informal argument is that a sentence is transportable if it considers the elements of the principal base sets as points without structure. We should not confuse this criteria or any formalization of this criteria with Marshall et al. theorem. For example, stating $0\in 1$ is making a statement about the structure of an element of the base set. I agree that this is not transportable, but I don't see the need to make a link with type theory. In fact, making such a link could lead to an incorrect generalization. In this perspective, the question is whether the truth of the sentence $\exists x\in P(X)\;\exists y\in X\times X\;\;x\in y$, with your modified definition of pairs, depends on the structure of the elements of the base sets. I believe that it does not, but I have to think about that. It requires a formal treatment. Dominic Mayers (talk) 15:03, 28 March 2018 (UTC)

OK, I agree with the "points without structure" idea. (About Marshal et al, true, I did not check it, but on the other hand, for now we did not see any counterexample.) Now, if the points of X are structureless, and $x\in P(X),$ and $y\in P(P(P(X))),$ and nevertheless $x\in y,$ then $x\in P(P(X))\cap P(X);$ well, it really can happen when x is the empty set! But only in this case, since an element of x (if any) must be both a point and a set of points. (Or may a structureless point be equal to the empty set?? I vaguely recall that in some set theory atoms are single-element sets that belong to themselves.) Now the question is, whether a pair may be (implemented as) the empty set. Well, probably it can for some implementation of pairs; but not for "my" implementation. Boris Tsirelson (talk) 16:13, 28 March 2018 (UTC)

Not sure what is the conclusion. If it is always false, it is transportable. What is the conclusion? Note that, if it is always false, it is equivalent to a sentence of type theory. So, it does not contradict Marshall et al. The "equivalent" in the statement of the theorem is important ! Dominic Mayers (talk) 16:21, 28 March 2018 (UTC)

Still thinking, but just want to confirm that, if we expand the definitions of pairs, we have that you agree that $$\exists x\in P(X)\,\exists x_{1},x_{2}\in X\,:\,(x=\{x_{1}\}\vee x=\{x_{1},x_{2}\})$$ is transportable, but you say that $$\exists x\in P(X)\;\exists x_{1},x_{2}\in X\;:\;x=\{\{x_{1}\},\{x_{1},x_{2}\}\}$$ is not transportable. Dominic Mayers (talk) 16:21, 28 March 2018 (UTC)

Yes I say it. The former condition is transportable, since it holds for all nonempty X. Now consider the latter. Clearly, some sets X violate this condition. (For example, this happens when all elements of X are sets of more than one element.) However, the set of all finite ordinals satisfies this condition: take $x_{1}=0=\emptyset$ and $x_{2}=1=\{0\},$ then $\{\{x_{1}\},\{x_{1},x_{2}\}\}=\{\{0\},\{0,1\}\}=\{1,2\}\in P(X).$ (Or simpler: $x_{1}=x_{2}=0.$) And by the way, the set of all finite ordinals greater than 1 violates this condition (since it contains only sets of more than one element). Here you see a very simple isomorphism of Peano models ($n\mapsto n+2$) that breaks the transport. Boris Tsirelson (talk) 17:07, 28 March 2018 (UTC)

Yes, simply $X=\{a,b,\{a\},\{a,b\}\}$ and $X'=\{a,b,c,d\}$ is enough. It hold with $X$, but not with $X'$. My assumption that the canonical extension on the universe of sets was a bijection when there is a global bijection on the principal base sets was wrong. This is why the idea for the simple proof fails. Dominic Mayers (talk) 17:47, 28 March 2018 (UTC)

"...today you require (in addition) $X_{1}\cap X_{2}=\emptyset ;$ tomorrow you'll require (in addition) $X_{1}\cap (X_{2}\times X_{3})=\emptyset ;$ then also $X_{1}\cap (X_{1}\times X_{1})=\emptyset ,$ $X_{1}\cap P(X_{1})=\emptyset$ etc. Then you'll observe that finite ordinals are no more an implementation of natural numbers." (Quoted from User talk:Tsirel#Definition of base sets.)   :-)   Boris Tsirelson (talk) 18:04, 28 March 2018 (UTC)

There is no connection, because I was looking for constraints on languages used to define species of structures, not for a new mathematical framework to replace set theory. Also, it is still true that I can modify slightly the definition of transportable languages so that every sentence in a transportable language is transportable. I only have to require that only variables relativised to $P(V_{n})$, where $V_{n}$ is the universe of sets, occur on the right of $\in$. So, the statement was basically correct, except that now, thanks to you, I see that there might not exist any "natural" transportable language, even for weak transportability. Yes, in order for all the sentences of a language to be transportable, it seems that we need that, for each occurrence of $x\in y$, the types are respected, because this guarantees that $x\in y$ is equivalent to $F(x)\in F(y)$, where $F$ is the canonical extension. I understand that better now. Dominic Mayers (talk) 18:41, 28 March 2018 (UTC)

I often fail to follow your ideas, but if indeed "the types are respected", then everything should be OK with me. Boris Tsirelson (talk) 18:49, 28 March 2018 (UTC)

The difference between me and you is that I wanted to see the need for the types to be respected only from the point of view of transportability. Marshall et al. was not convincing because we could not see the basic idea and they do use assumptions. By the way, one of these assumptions is that there are no auxiliary sets. Maybe it's not so bad, because in most cases we could make these auxiliary sets principal sets. Still, it is puzzling that there is this assumption. Dominic Mayers (talk) 18:55, 28 March 2018 (UTC)

You said that what is missing is a way to explain it to a larger audience. Here is what I feel now was missing. The proof for the obvious side (every sentence of type theory is transportable) should be given. The one that I have in mind would use $x\in y$ is equivalent to $F(x)\in F(y)$ when the types are respected, because $F$ is a bijection. Now, from this starting point, one can informally argue that, if instead we have $x\in y$ where the types are not respected, then $F$ is not a bijection on the larger domain that includes both $x$ and the elements of the set $y$, and thus the argument fails. This has the benefit of providing some ideas why types must be respected for transportability, without arguing in terms of analogy with computer science, evilness, etc. Your examples, etc. helped me a lot. On the other hand, your discussions in terms of computer science, etc. were perhaps good ways to preach for type theory, but that did not help me to understand why types had to be respected for transportability, Dominic Mayers (talk) 19:14, 28 March 2018 (UTC)

It's worst than not a bijection, it is not even well defined. If $X=\{a,\{a\}\}$, then $P(X)=\{\emptyset ,\{a\},\{\{a\}\},\{a,\{a\}\}\}$. The canonical extension of $f$ on $P(X)$ evaluated at $\{a\}$ is $\{f(a)\}$, which is usually not $f(\{a\})$. So, what is $F(\{a\})$? Dominic Mayers (talk) 20:01, 28 March 2018 (UTC)

I could not be helpful in this aspect, since I never felt any other driving force behind transportability, only the type control. And so, I did not dream about a much simpler approach. Boris Tsirelson (talk) 20:07, 28 March 2018 (UTC)

Vaguely I'd say that structureless points behave like free generators (of a group or something alike). Thus, we deal effectively with the scale of sets as if the points are structureless. This is possible exactly due to type control. Boris Tsirelson (talk) 20:13, 28 March 2018 (UTC)

$F(\{a\})$ is $\{f(a)\},$ and $F(\{\{a\}\})$ is of course $\{f(\{a\})\},$ just because $F(\{x\})$ is $\{f(x)\}$ for all x. It does not matter whether or not this x is entangled with some a by a relation visible only when internal structure of points is taken into account (instead of being hidden), which is evil! Just avoid evil, and be happy. Your X is $\{a,b\};$ the relation $b=\{a\}$ holds accidentally, but is of no consequences (as long as we avoid evil and are happy). Boris Tsirelson (talk) 20:20, 28 March 2018 (UTC)

The canonical extension $F$ of $f$ evaluated on the base set $X$ is $f$, by definition. The canonical extension on $P(X)$ is (in a sloppy way) $F(\{x_{i}\})=\{f(x_{i})\}$. If it turns out that $\{x_{i}\}$ is also an element of $X$, then we have two definitions: $F(\{x_{i}\})=f(\{x_{i}\})$, when $\{x_{i}\}$ is seen in $X$ and $F(\{x_{i}\})=\{f(x_{i})\}$, when $\{x_{i}\}$ is seen in $P(X)$. These two can be different. I know, within type theory this is not an issue, because $X$ and $P(X)$ are different types. However, I was just pointing out why my argument was wrong when it used a well defined canonical extension on the universe of sets. What you say about free generators seems interesting. Do you know if it can be made more rigorous? Dominic Mayers (talk) 20:52, 28 March 2018 (UTC)

As for me, it just means that relations between the points (implemented as sets instead of being structureless), if any, have no consequences (for me, since I respect types). The corresponding rigorous statement is the transportability of all sentences that respect types. That is, the scale of sets, endowed with all structure that grows from the given structure on base sets, is isomorphic to what it would be over structureless points. But, I know, you seek a different answer. Boris Tsirelson (talk) 07:24, 29 March 2018 (UTC)

Again, you repeat a few times the rule "avoid evil", which means "respect types". But, again, my interest was in understanding why we have to follow this rule for transportability and see, whether we can have a different rule, etc. Just knowing the rule was not sufficient. Dominic Mayers (talk) 21:01, 28 March 2018 (UTC)

_{I could not be helpful in this aspect, since I never felt any other driving force behind transportability, only the type control. And so, I did not dream about a much simpler approach.} Before that you wrote that you were interested in explaining to a larger audience. When we take both together, does it mean that you wanted to preach it as if it was a religion? Seriously, it is not so obvious that respecting types is fundamental for transportability and reducing this fact to a notion of "evil" or an analogy with programming languages is missing an occasion to discover deeper principles that explain why. At first, I thought it was a consequence of no global bijection in the definition of isomorphism. Had I been right, it would not have been a deep thing, but now that I see that it remains important, even if we have a global bijection, then I see, ok, it's fundamental. There might be some kind of general principle that explains that. For now, simply this informal argument based on the proof of the easy side, is better than nothing. Do you appreciate what I mean? Dominic Mayers (talk) 21:39, 28 March 2018 (UTC)

Yes, more or less, I start to appreciate. After trying to throw away the type check, you intend to examine thoroughly many attempts to weaken the type check here and there. Either you'll discover a new theory beyond the existing one, or you'll convince yourself (and then, others) that the type check is the only way. That is OK; I no longer prevent you from trying (nor participate). But, back to Wikipedia, I have a question.

Imagine that someone visits the article Field (mathematics), delists it from good articles "because it just introduces the 6 postulates as if it was a religion", tries to throw them away, unsuccessfully, and then starts to examine thoroughly many attempts to weaken the 6 postulates here and there. What will you say about his activity? "Shoot First... Ask Questions Later"? Boris Tsirelson (talk) 05:30, 29 March 2018 (UTC)

You confuse three statements: the consistency of type theory, the possibility to base all mathematics on type theory and the more specific statement that type checks is necessary and sufficient for transportability. I never argued against the first statement. I perhaps expressed an opinion on the second statement, but that would be because of your interest on that issue, not mine. My main focus is only the third statement. I don't think that it makes sense to compare this statement with the article Field (mathematics). It is totally reasonable to question whether type check is necessary and sufficient for transportability, especially when the question regards the hidden assumptions in the statement, whereas questioning the interest of the postulates of field theory is not. You are not being fare.

Perhaps you should read Marshall et al., because you seem to believe blindly on something you don't really understand. You don't realize that there are a lot of assumptions in Marshall et al. The first one that I mentioned is the choice of the constants in the language that define the possible sentences. Marshall et al. do not prove that amongst all possible sentences, only the sentences of type theory are transportable. This is not at all what they prove. They prove that amongst the sentences of a given language ${\cal {L}}$, those and only those sentences that are equivalent to a sentence of type theory are transportable. Therefore, it is much weaker than you think. If we interpret their result, as you do, to mean that only the sentences that are equivalent to a sentence of type theory is transportable, then the definition of the language ${\cal {L}}$ is a strong assumption in that interpretation. The only thing that I do is try to explore what happens when we remove or change this strong assumption, which is a completely natural and healthy mathematical attitude. Dominic Mayers (talk) 11:31, 29 March 2018 (UTC)

Cyclic reasoning in all arguments to support type checks for transportability

In all arguments that I see to support type checks as a criteria for transportability, I see cyclic reasoning. Certainly, I see it in Marshall et al. Types are almost built-in the language ${\cal {L}}$. It is almost surprising that their proof is complicated, given that they almost have the conclusion in their hypothesis. Also, the analogy with programming languages is also cyclic reasoning, because most likely type theory has influenced programming languages. Then, we use an analogy with programming languages to support type theory. It could be that we are just lacking imagination and different constructions than types would do better. Therefore, the rigorous support for type checks is basically inexistent. Again, I am not interested in the general question, but only in the fundamental question of transportability. I think it is good not to force ourselves within type theory, if we want to discover general principles behind transportability. Transportability is way too much a fundamental concept to be forced within a specific theory. Your feeling that type theory is the answer is theoretically empty and based on cyclic reasoning. Dominic Mayers (talk) 11:57, 29 March 2018 (UTC)

On the other hand, the appreciation of any set of postulates and thus of any theory based on postulates is usually based on some form of cyclic process. So, ranting against such a cyclic process is tantamount of ranting against the natural way to appreciate a theory. Fine, it is Ok that there is a cyclic process in the way we appreciate type theory. However, we have to have the modesty of recognizing the limitation of such a cyclic process and recognize that the most fundamental concepts such as transportability must be analyzed outside any such specific set of postulates. Dominic Mayers (talk) 12:23, 29 March 2018 (UTC)

_{That is, the scale of sets, endowed with all structure that grows from the given structure on base sets, is isomorphic to what it would be over structureless points. But, I know, you seek a different answer.} No, I am not looking for a different answer. I would be very happy with this answer. You might believe that it is the answer that Marshall et al. has provided. It is not. Again, there are a lot of assumptions in Marshall et al. On the contrary, I would be delighted to have this answer, that is, to transform this conjecture into a theorem. Often, when you face a conjecture, you try to disprove it. It does not mean that you don't want it to be true. It's just a natural way to analyze the conjecture. The difference between me and you is that you don't realize or don't want to deal with the fact that it is a conjecture. Dominic Mayers (talk) 12:45, 29 March 2018 (UTC)

References

Disambiguation pages

The purpose of a Wikipedia disambiguation pages, such as David Kaye, is to assist readers in finding the most relevant article on a similarly named topic. As there is no target article for the David Kaye Dean of Research, the entry should not appear on the disambiguation page. If an article is created in the future then it would be appropriate to include the entry.--Jezebel's Ponyo^{bons mots} 21:11, 5 June 2018 (UTC)

@Ponyo : The problem with this restriction is that when you edit an article on a topic and someone notable made a contribution to the subject, you naturally want to explicitly name the person. Because you name the person in the article, you want to disambiguate the name. Yet, the specific contribution for which you name it is not so important. I mean, it is very important in the article, but not important in comparison to the notability of the person. You are not an expert about this person. You will not want to create an Wikipedia about this person, even though it is a notable person. Yet, you want to name the person, because you can easily see that he is a notable person: you don't need to be an expert about the person to see that. With the rule that you mention, the only other option would be to mention in the disambiguation page the article where you name the person. But, this is not so adequate either, because this is far from being the reason why the person is notable. So, this rule kind of force Wikipedia contributors to not name a notable person with an ambiguous name, if there is no Wikipedia article about this person. It's an all or nothing situation.Dominic Mayers (talk) 22:23, 5 June 2018 (UTC)

@Ponyo The short version is I would understand that only notable persons or entities should be mentioned in a disambiguation page and perhaps the notability of the person should be explained in the talk page, but it seems a complicated and useless restriction to require that you prove the notability of this entity by an article in Wikipedia. It is useful, even important, to disambiguate between any notable identity with this name, irrespective of whether there is an article about this identity in Wikipedia. The argument that, if the identity is notable, then we should have a Wikipedia article about it is an all or nothing argument. I don't see what is gained. I hope the idea is not to force editors to write articles about every notable entity that they need to name. Dominic Mayers (talk) 22:40, 5 June 2018 (UTC)

I don't think you understand. Disambiguation pages serve a very specific purpose on Wikipedia; they help readers find the article they are looking for. It's not a list of notable and possibly notable people with the same name, it's a navigational aid to help readers find a specific article. Adding non-navigable links to such pages would be like restaurants putting meals on their menu that they don't actually sell. --Jezebel's Ponyo^{bons mots} 22:51, 5 June 2018 (UTC)

Actually, I understood very well that this rule assumes that a disambiguation page should have this specific purpose that you describe. I understood very well that you consider that this should be the only purpose of the disambiguation page. However, if you think about it, you can easily see that another purpose is also very important: making sure that a person does not go to a wikipedia article that is about a different entity with the same name, irrespective of whether this other identity is in the Wikipedia menu. To continue with your analogy, if a Londonian restaurant in the USA offers some liquor with the pasta, it would make sense that it disambiguates it from the USA liquor, even though the USA liquor is not in its menu (say it does not have a license for that). Dominic Mayers (talk) 23:20, 5 June 2018 (UTC)

You were right when your wrote "It's an all or nothing situation" above. You want disambiguation pages to serve a purpose other than what they were designed for. You can try to get consensus for your vision of how DAB pages should work by beginning a discussion at Wikipedia talk:Manual of Style/Disambiguation pages, but unless there is consensus for your proposed changes, you will need to abide by the current system Wikipedia uses for article navigation.--Jezebel's Ponyo^{bons mots} 23:27, 5 June 2018 (UTC)

No, I will not go that far. I just wanted to say that, from the point of view of the waiter at the restaurant, it would be useful to have a disambiguation in the menu (say about liquor) even if one of the items with the same name is not served. Still, I do understand that they will not change the menu for that reason, because it is costly. The analogy works well. It is also costly in time to verify that an entity is notable. I can see that they would want to do that only if it will result in an article. To put it in another way, it might not be reasonable to expect that checking that an entity is notable in a talk page will be done properly, because of lack of human resources. Dominic Mayers (talk) 00:31, 6 June 2018 (UTC)

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ArbCom 2018 election voter message

Hello, Dominic Mayers. Voting in the 2018 Arbitration Committee elections is now open until 23.59 on Sunday, 3 December. All users who registered an account before Sunday, 28 October 2018, made at least 150 mainspace edits before Thursday, 1 November 2018 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

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ArbCom 2019 election voter message

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June 2020

Please refrain from making unconstructive edits to Wikipedia, as you did at Talk:Falsifiability. Your edits appear to constitute vandalism and have been reverted. If you would like to experiment, please use the sandbox. Repeated vandalism may result in the loss of editing privileges. Thank you. Drmies (talk) 02:02, 26 June 2020 (UTC)

You may be blocked from editing without further warning the next time you vandalize Wikipedia, as you did at Talk:Falsifiability. Drmies (talk) 02:07, 26 June 2020 (UTC)

Please, do not attempt to "own" the article Falsifiability and let other editors decide what to include as well, not only yourself

Considering your extensive number of hundred or even thousands of edits on Falsifiability, and the fact that you usually delete others' editions on that page, I invite you to read Wikipedia:Ownership of content, where it is stated that "All Wikipedia content—articles, categories, templates, and other types of pages—is edited collaboratively. No one, no matter how skilled, or how standing in the community, has the right to act as though they are the owner of a particular page". James343e (talk) 17:30, 1 July 2020 (UTC)

You should not construe the great amount of well sourced content that I added to this page as an intent to "own" the page. The fact that you do that hurts me, because I think the content that I added was, on the contrary, very useful and should be simply appreciated.

Context that should concern @Tim bates, Sunrise, and Drmies: Here is the version in which I left it. I am not responsible for non sense such as "refutability is the capacity for a statement, theory or hypothesis to be contradicted by evidence" that were added later: It's so weird. Normally, evidence refers to something that is validated to some degree, usually by statistical methods, meta analysis, etc. For example, this is what we expect from evidence in the expression "evidence-based medicine". Given that this is the usual meaning of the term evidence, can the purported observation of an apple that moves from the ground up to a branch and then starts to dance from one branch to another in the tree be evidence for anything? No! But this is exactly the example provided by Popper to show that Newton's theory is falsifiable. Popper said so many times that falsifiability has nothing to do with facts. Popper, Keuth, Thornton and many others explain that it has something to do with technology (e.g. to observe the apple), but it has nothing to do with facts or evidence (as clearly illustrated by the dancing apple).

Of course, one can extend the meaning of the term evidence to include dancing apples. The required extension is simply that any sequence of observations that are separately possible can be used as evidence, but it is a pious hope to expect that the readers will correctly guess this new meaning of "evidence". It's hard to tell how readers understand the sentence above. Most likely, it is interpreted using the understanding or feeling that it is almost impossible that we will discover a scientific law that cannot ever be falsified. The feeling is that, if it really cannot be falsified by (normal) evidence, then it's most likely not scientific. It's a very natural interpretation that uses the usual meaning of both "falsifiability" and "evidence". It makes sense, but as pointed out by Thornton, Popper emphasized that this interpretation faces all the problems of falsification that (technical) falsifiability avoid. Nevertheless, many people might feel good about this natural, but problematic interpretation. There is a confusion here between making people feel good about their current incorrect expectation and conveying a correct new basic idea in simple terms. Only the former, not the latter, is most likely accomplished by the sentence above. The previous version that I provided matched perfectly with the content of the article, which was verifiable in good quality primary and secondary sources. The quoted sentence above presents a different view point. It's not a simpler or intuitive version of it. On the contrary, Popper insisted that we distinguish between these two meanings. So sad, so sad.

My reply to @James343e: Since when the fact that an editor is a useful contributor to a page and becomes the main contributor to the page means that he thinks that he owns the page? Certainly, it was not the case with me. I obviously understood very very well that I do not own the page and was always extremely aware that the opinions of other editors would affect the content of the page and that consensus was very important. In fact, I decided to stop working on the article because of a lack of consensus about what the sources say and I have not even considered a second that my view should define this consensus and that others should quit editing if they don't agree with me. I could not possibly think in that way, because I know too much how forums work, not just Wikipedia.

You also in a completely biased manner point toward the fact that I deleted some content written by others. It's biased because you did not consider all the content that was added by others and which I kept. (I did try to remove calumnious statements about me in the talk page and I feel I should have been allowed to do so.) The reality is that I mostly kept the content that was added by others in the article and whenever I removed some content, anyone was free to add back the content and I never entered in an edit war in an article. I removed content in a normal manner (more often modified the content by adding sourced material to it). For example, Banno has deleted, I believe, a lot more content written by others than I did and, if I were to use your criterion, I would have to conclude that he acted, at the time, as if he owned the article.

Finally, I lost access to my account and thus can only make a few anonymous edits. I do not intend to do a lot of these anonymous edits. In fact, given the attitude that you show, I don't think that I will do any edits in the future. Dominic Mayers (talk) 12:18, 6 July 2020 (UTC) (signature copied and edited by hand, because I lost access to my account). — Preceding unsigned comment added by 173.206.162.45 (talk)

Well, Dominic Mayers, I assume then that we will see no more edits in the article. I note also that the above messages took thirty edits to produce, and that's one of the problems in the article. Drmies (talk) 00:47, 10 July 2020 (UTC)

@Drmies: I am not perfect. For reasons that I cannot explain, the option to wait longer before publishing does not work with me. Instead of criticizing aspects that can be deal with (by doing a diff that includes all the edits), you should appreciate the new, pertinent and well sourced material that I added in the article. This negative attitude hurts me. Dominic Mayers (talk) 15:01, 10 July 2020 (UTC)

ArbCom 2020 Elections voter message

Hello! Voting in the 2020 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 7 December 2020. All eligible users are allowed to vote. Users with alternate accounts may only vote once. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate in the 2020 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 02:51, 24 November 2020 (UTC)

Nomination for deletion of Template:Highlighted anchor

Template:Highlighted anchor has been nominated for deletion. You are invited to comment on the discussion at the entry on the Templates for discussion page. – Jonesey95 (talk) 17:19, 17 May 2022 (UTC)