It is my first time that I met the word "entails". In mathematical texts, one usually sees "if and only if", "implies" or "iff" which bear no ambiguity. In the following definition:
Does "entails" mean "implies"? It certainly seems so. Also In one PHD thesis I am reading, the author quotes this definition using the word "implies" instead of the word "entails". So far so good, but next comes a corollary:
But if "entails" means "implies" then this corollary is not true. Since function
$$ f_1 \left( x \right) := x $$
clearly supports
$$ f_2 \left( x \right) := 1 $$
according to the definition above, but not the other way around.
This is because $x' \geq x \Rightarrow 1 \geq 1 $ but $ 1 \geq 1 $ does not imply $x' \geq x $ for all $x', x \in \mathbb{R}$.
What am I missing here?
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$\begingroup$I'm 100% British (English, even), and I think this is strange and confusing usage.
According to my (American) thesaurus, some synonyms of "entails" are "brings about", "calls for", "demands", "causes", "gives rise to", "leads to", "necessitates", "requires".
From a logic/mathematics point of view, some of these terms mean "if" and some mean "only if".
I have never seen "entails" used to mean "equivalent to", and the thesaurus doesn't list this as a synonym, but who knows what these authors had in mind. Pretty poor writing, in my opinion.
Clarification of Synonyms
Let's split the terms into two groups:
I think it's clear that "brings about", "causes", "gives rise to", "leads to" all mean the same thing. Let's represent this group by "causes".
Similarly, I think "demands", "necessitates", "requires" all mean the same thing. Let's represent this group by "requires".
"A causes B" means that A implies B. In other words, "B is true if A is true".
"A requires B" means that "A can be true only if B is true"
So, in my view, the list of synonyms contains terms with two quite different meanings, which I (rather sloppily) characterised as "if" and "only if" terms.
$\endgroup$ 2 $\begingroup$You are perfectly right, and "entails" certainly should not be taken to mean "is equivalent to". I can just offer one possible explanation for this text: if $\geq$ is a total ordering and an "objective function" is always injective (so "ob" entails "in", I don't know the definition of "objective") then this relation is indeed symmetric. The function $f_2$ in your counterexample is of course not injective.
For in this case, assuming $f_i\uparrow f_j$ and $f_j(x')\geq f_j(x)$, one either has $x'=x$ and therefore $f_i(x')=f_i(x)$, or (by injectivity) $f_j(x')>f_j(x)$, whence (by $f_i\uparrow f_j$) it cannot be that $f_i(x')\leq f_i(x)$, so (by the total ordering) $f_i(x')>f_i(x)$; in summary we get $f_i(x')\geq f_i(x)$ in both cases.
I should add that "$X=\Bbb R^n$" in the citation makes it rather unlikely that objective functions are supposed to be injective. If this is about objective functions in optimization problems, which take their values in $\Bbb R$, then it is almost certainly wrong, as being injective is incompatible with being continuous for $n\geq2$. And for any non-injective objective functions you can do as you did in the counterexample.
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