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I know what sufficient resp. necessary means, but I'm confused, when our professor uses the terminology weaker resp. stronger. I couldn't yet find out, what the translation of these pair of words to the pair necessary stronger is (at least I'm guessing that A being weaker/stronger than B is just a different way to say that A is sufficient/necessary for B) ?

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Also relevant: math.stackexchange.com/questions/53708/… –  Asaf Karagila Oct 18 '12 at 14:44

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up vote 5 down vote accepted

I can think of two relevant ways in which weaker and stronger are used. A statement $A$ is stronger than a statement $B$ if $A$ implies $B$; $A$ is weaker than $B$ if $B$ implies $A$. Equivalently, $A$ is stronger than $B$ if $A$ is sufficient for $B$, and $A$ is weaker than $B$ if $A$ is necessary for $B$.

We also speak of one theorem being stronger or weaker than another. Consider the theorems $A\Rightarrow B$ and $A'\Rightarrow B\,'$. If the hypothesis $A$ is weaker than the hypothesis $A'$, or the conclusion $B$ is stronger than the conclusion $B\,'$ (or both), the theorem $A\Rightarrow B$ is stronger than the theorem $A'\Rightarrow B\,'$ (and of course $A'\Rightarrow B\,'$ is weaker than $A\Rightarrow B$).

Here’s an example of this usage. Consider the following two theorems.

Theorem 1. Let $X$ be a complete metric space, and suppose that $\{G_n:n\in\Bbb N\}$ is a family of dense open subsets of $X$; then $\bigcap_{n\in\Bbb N}G_n\ne\varnothing$.

Theorem 2. Let $X$ be a complete metric space, and suppose that $\{G_n:n\in\Bbb N\}$ is a family of dense open subsets of $X$; then $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$.

Since a dense subset of $X$ must be non-empty, the conclusion of Theorem 2 implies the conclusion of Theorem 1: the conclusion $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$ is stronger than the conclusion $\bigcap_{n\in\Bbb N}G_n\ne\varnothing$. This means that Theorem 2 is stronger than Theorem 1: it reaches a stronger conclusion from the same hypothesis. In a sense it says more.

Here we strengthened Theorem 1 by strengthening its conclusion. One can also strengthen a result by using a weaker hypothesis to reach the same conclusion; here again the outcome is a theorem that in a sense says more, because it uses less information to reach the same conclusion.

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Shouldn't the hypothesis be weaker and conclusion stronger ? Since if, e.g, only the hypothesis of the first theorem is weaker then the hypothesis of the second, we might have two theorems which actually are uncomparable: Consider $A$="$x$ is an prime integer", $A'=$"$x$ is an integer", $B=$"$x$ isn't $4$" and $B'=$"$x$ isn't $\sqrt(2)$". Thus, although $A$ is weaker than $A'$, i.e. $A'$ implies $A$, we can't conclude that $A \Rightarrow B$ is stronger than $A' \Rightarrow B'$, i.e. $ (A \Rightarrow B) \Rightarrow (A' \Rightarrow B)'$, since neither $B$ implies $B'$ nor $B'$ implies $B$. –  temo Oct 26 '12 at 17:49
    
@temo: The example is irrelevant: your $B$ and $B'$ are not comparable, and I said nothing about such comparisons. The answer to your first question is no: if the hypothesis of Th. 3 is weaker than that of Th. 4, and the conclusions of the theorems are identical, then Th. 3 is stronger than Th. 4. –  Brian M. Scott Oct 26 '12 at 18:31
    
Well, ok, but isn't it somewhat annoying to have the words "weaker"/"stronger" mean a different thing in the case of theorems that in the case of statements ? Since, after all, a theorem is a statement, so it would seem nice if "weaker"/"stronger" for theorems were just a special case of "weaker"/"stronger" for statements - or at least, if "weaker"/"stronger" for theorems would imply "weaker"/"stronger" for statements (the latter being the case only if one defines "weaker"/"stronger" for theorems as in my question above: hypothesis be weaker and conclusion stronger) ? –  temo Oct 27 '12 at 14:48
    
@temo: I’m not sure what, but you’re definitely misunderstanding something, because weaker/stronger for theorems is a special case of weaker/stronger for statements in general. In the example in my previous comment, Th. 3 makes a stronger statement than Th. 4, even though they have the same conclusion, because it says that that conclusion follows from less. –  Brian M. Scott Oct 27 '12 at 14:53
    
Ah, I think I understood where our differences came from: In your answer you said "If the hypothesis $A$ is weaker than the hypothesis $A′$, or the conclusion $B$ is stronger than the conclusion $B′$ (or both), [...]" but you didn't say that in the first case the conclusion has to stay identical or be stronger, resp. in the second case the hypothesis has to be the same or weaker, so I didn't assume that that had to be the case, i.e. to call the theorem "stronger" the hypothesis could be weaker but the conclusion could be anything - which [...] –  temo Oct 27 '12 at 16:19

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