# Sufficient/necessary vs. weaker stronger

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) ?

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.

• 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
• @temo: That would explain the misunderstanding, yes. I was thinking specifically of what it means to strengthen a theorem of the form $\varphi\to\psi$: either you weaken $\varphi$, or you strengthen $\psi$ (or of course both). I wasn’t thinking of comparing two unrelated theorems drawn from a hat, so to speak. Sorry not to have made that clearer; for some reason that possible misinterpretation just didn’t occur to me. – Brian M. Scott Oct 27 '12 at 16:28