What are some examples of logically weak (when there are more objects satisfying) but being called 'strong' or logically strong (when there are fewer objects satisfying) being called 'weak'?
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5$\begingroup$ There are two partial orders on a Coxeter group $W$, often called the weak Bruhat order and the strong Bruhat order. Every relation in the weak order is a relation in the strong order, but not conversely. So, to say $u \leq v$ in weak order is a stronger statement than to say $u \leq v$ in strong order. However, note that nowadays it is more common to refer to "weak Bruhat order" as just "weak order" and "strong Bruhat order" as just "Bruhat order." $\endgroup$– Sam HopkinsMar 17, 2023 at 2:10
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8$\begingroup$ I always tell myself the strong Bruhat order is stronger in the architectural sense, as the extra edges in the Hasse diagram hold it together better. $\endgroup$– darij grinbergMar 17, 2023 at 2:14
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4$\begingroup$ @SamHopkins Interesting; I have never found this terminology confusing. It seems natural to me that a "partial" ordering is "weaker" than a "total" ordering. So the closer you are to a total ordering, the stronger it is, and the closer you are to an antichain, the weaker the ordering is. $\endgroup$– Timothy ChowMar 17, 2023 at 13:00
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5$\begingroup$ Strong induction is stronger than weak induction because it has the same conclusion, but weaker hypothesis. So what makes you say that it is a “weaker condition” than weak induction? (What do you mean by “condition” here, anyway?) $\endgroup$– Emil JeřábekMar 17, 2023 at 13:55
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2$\begingroup$ ... theory such as Robinson’s arithmetic Q (or better, $\mathrm{PA}^-$). Over such a background theory: (1) strong induction for any particular formula $P$ implies weak induction for the same formula, but (2) in general, weak induction for $P$ does not imply strong induction for $P$. Thus strong induction is strictly stronger than weak induction in this sense. However, strong induction for $P$ follows from weak induction for a sutable different formula, hence if you postulate both schemata for all formulas, they are equivalent. Now, what happens if you drop the background theory? ... $\endgroup$– Emil JeřábekMar 18, 2023 at 7:03
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5 Answers
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Munkres's book Topology (p. 78) contains the following warning about coarser and finer topologies:
Many mathematicians use the words "weaker" and "stronger" in this context. Unfortunately, some of them (particularly analysts) are apt to say that $\mathcal T'$ is stronger than $\mathcal T$ if $\mathcal T' \supset \mathcal T$, while others (particularly topologists) are apt to say that $\mathcal T'$ is weaker than $\mathcal T$ in the same situation! If you run across the terms "strong topology" or "weak topology" in some book, you will have to decide from the context which inclusion is meant. We shall not use these terms in this book.
For instance, the 'weak topology' in the definition of a CW complex is the finest topology for which all cell inclusions are continuous. But weak topologies on a topological vector space $V$ are usually the coarsest topology for which certain maps $V \to \mathbf R$ are continuous.
The difference here comes from whether you think a topology is weaker if it has more opens or if more sequences converge.
(Like Munkres, I have adapted the convention to stick to the unambiguous terms coarser and finer.)
answered Mar 17, 2023 at 14:25
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2$\begingroup$ 'Weak topology' occurs when the situation is at the lower edge of continuity (with the minimal number of continuous maps): For a topological space $X$, the finer the topology on $X$ is, the less possibility of continuity a map $f$ with target $X$ will have; Conversely, the coarser the topology on $X$ is, the less possibility of continuity a map $f$ with source $X$ will have. Similarly, 'strong topology' occurs when the situation is at the upper edge of continuity. In this sense, the use of 'strong' and 'weak' is subtle, but not quite 'confused'. $\endgroup$– ZeroxMar 17, 2023 at 19:05
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1$\begingroup$ but I wonder how many people would really call "stronger" a coarser topology. My guess is very few. I think the situation is not symmetric. $\endgroup$ Mar 17, 2023 at 19:56
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I believe that strong monoidal functors were originally called "weak monoidal functors". The newer name indicates that it's a stronger condition than being a lax monoidal functor, while the original name indicates that it's a weaker condition than being a strict monoidal functor.
answered Mar 17, 2023 at 18:41
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There are certainly examples where the terms "strong" and "weak" can be confusing, though I'm not sure I would call any of them "mistakes" per se. Even in your example of "strong induction" versus "weak induction," I am not sure what you mean when you say that "strong induction is a weaker condition." Are you saying that if an induction is weak then it is also strong, but not necessarily vice versa? That doesn't seem to be true, at least not according to how I've heard the terms "strong induction" and "weak induction" used.
But the example that comes to mind is the "weak law of large numbers" and the "strong law of large numbers." The strong law is so called because its conclusion (almost sure convergence) is stronger than the conclusion of the weak law (convergence in probability). However, the terminology can give the impression that the weak law is a straightforward corollary of the strong law, since that's what we normally mean by saying that Theorem A is "stronger" than Theorem B. But the weak law does not follow straightforwardly from the strong law, because the hypotheses of the weak law are different.
The type of example alluded to by Sam Hopkins in a comment also arises frequently. A structure on a set may be called "weaker" than another if there is less structure. So for example, a topology $\mathscr{U}$ on a set $X$ is weaker than a topology $\mathscr{V}$ on $X$ if $\mathscr{U} \subseteq \mathscr{V}$. But this means that for a given subset $U \subseteq X$, the condition $U\in \mathscr{U}$ is stronger than the condition $U\in \mathscr{V}$. I wouldn't call this an "error" but some people might find it confusing.
answered Mar 17, 2023 at 13:34
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$\begingroup$ In the topology example, I thought the usual terms were "finer" and "coarser"? EDIT: Nevermind, you are right, "stronger" and "weaker" are also used for this... $\endgroup$ Mar 17, 2023 at 13:35
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$\begingroup$ @SamHopkins That terminology is used, but certainly "weak" is used as well. Certainly the terminology "weak-* topology" is pretty entrenched; I've never heard the Banach-Alaoglu theorem ever stated without using the term "weak-* topology." $\endgroup$ Mar 17, 2023 at 13:38
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$\begingroup$ In what sense are the hypotheses of the weak law of large numbers different to those of the strong law of large numbers? Obviously, one can come up with situations where the conclusion of the weak law holds and that of the strong law doesn't. But aren't the names both generally understood to refer to the case of a sequence of i.i.d. random variables with finite mean? $\endgroup$ Mar 17, 2023 at 17:29
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$\begingroup$ @JamesMartin Wikipedia lists a few examples where the weak law holds but the strong law doesn't. In particular, finite mean isn't always necessary for the weak law to hold. $\endgroup$ Mar 17, 2023 at 21:49
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1$\begingroup$ Maybe it would have been clearer if I had said that the weak law does not always follow straightforwardly from the strong law, because the hypotheses can be different? In any case, the point is that it can be confusing: if the weak law is just an easy corollary of the strong law, then why do people bother talking about the weak law? $\endgroup$ Mar 18, 2023 at 1:54
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As an example where the adjectives "weak" and "strong" refer to hypotheses, rather than conclusions, a favorite of mine is (Gelfand-Pettis) "weak" vector-valued integrals, as juxtaposed to (Bochner) "strong" vector-valued integrals.
Apart from other hypotheses, such as that the vector space is locally convex and quasi-complete, the "weak" integral $\int_X f(x)\,d\mu(x)$ if a $V$-valued function $f$ is characterized by $$ \lambda\Big(\int_X f(x)\,d\mu(x)\Big) \;=\; \int_X \lambda(f(x))\,d\mu(x) $$ for all $\lambda$ in the continuous dual of $V$, noting that the right-hand side is just a scalar-valued integral. By Hahn-Banach, there is at most one such. Then, under various (useful-in-practice) hypotheses, one proves existence. To me, it's fairly amazing that the "weak" (because it refers to the dual) condition gives so much.
In contrast, the "strong" (Bochner) integral constructs an "integral", in analogy with Riemann and Lebesgue integrals, and then proves it has the desired properties (which would/do also follow from the "weak" condition!)
answered Mar 17, 2023 at 18:11
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$\begingroup$ But in this case isn't at least the logical strengths correct? In that a Bochner integrable function is Gelfand-Pettis integrable? $\endgroup$ Mar 17, 2023 at 19:12
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$\begingroup$ @WillieWong, ah, well, yes, the hypotheses's implications run in that direction, for sure. (Though some traditional formulations of Bochner integrals are too restrictive... I think Anton Deitmar showed that quasi-complete, locally convex, which is needed for "weak", also works for "strong"...) $\endgroup$ Mar 17, 2023 at 19:21
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It is sometimes said that the axioms of a group can be "weakened" by merely requiring the existence of a left identity element, and left inverses. A priori it seems that replacing the two-sided axioms with their one-sided counterparts could lead to more models. However, it turns out that in a semigroup with both a left identity and left inverses, the left identity must also be a right identity, and the left inverses are also right inverses. Therefore, the "weakened" axioms have exactly the same models as the usual group axioms, meaning that the term "weakened" is perhaps misleading. On the other hand, in a general algebraic structure, the existence of a left identity is still a strictly weaker condition than a two-sided identity. Thus, a collection of individually weaker axioms might still yield the same models as a collection of their individually stronger counterparts. This is an elementary observation, but an important one, since it reinforces how the terms "stronger" and "weaker" are sensitive to context.
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$\begingroup$ Perhaps more striking is that commutativity of addition is superfluous for commutative rings, but obviously very strong in general. $\endgroup$ Mar 17, 2023 at 22:02
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$\begingroup$ @Carl-FredrikNybergBrodda: Did you mean to say that it is superfluous for unital rings? $\endgroup$– JoeMar 17, 2023 at 23:57
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2$\begingroup$ Well, all my rings have an identity element (otherwise they are rngs). $\endgroup$ Mar 18, 2023 at 0:51
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2$\begingroup$ @Carl-FredrikNybergBrodda: Assuming that all rings are unital, I think that commutativity of addition is a superfluous axiom even for noncommutative rings. It follows by expanding the product $(1+1)(x+y)$ (see Bill Dubuque's answer here). $\endgroup$– JoeMar 18, 2023 at 1:01
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$\begingroup$ Yes, that was what I wanted to refer to — but I added an extra commutative in my comment. $\endgroup$ Mar 18, 2023 at 6:46