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I see (at least) two methods for proving the universally-quantified statement $$\forall a{\in} D~\big(P(x)\rightarrow Q(x)\big).$$

(1) Method 1

Let $a$ in $D$ (or, Let $a$ be any element in $D$).

$\cdots\cdots$ (some arguments)

$\therefore P(a)\rightarrow Q(a).$

Since we have made no special assumption on $a$, $P(a)\rightarrow Q(a)$ holds true for any $a$ in $D.$ Thus, we have shown that $\forall a{\in} D~\big(P(x)\rightarrow Q(x)\big).$

(2) Method 2

Arbitrary choose $a$ in $D.$

$\cdots\cdots$ (some arguments)

$\therefore P(a)\rightarrow Q(a)$

Since $a$ was arbitrarily chosen, $P(a)\rightarrow Q(a)$ holds true for any $a$ in $D.$ Thus, we have shown that $\forall a{\in} D~\big(P(x)\rightarrow Q(x)\big).$

Which formulation is more proper, logical and better structured? I think that Method 1 is an exact application of the Universal Generalisation rule, while Method 2 is somewhat not clear. Yes, we had indeed arbitrary chosen a element $a$ in $D,$ but someone may argue that "hey, it is because you're lucky that you randomly chose the right $a$ at the very beginning that makes the predicate true in the set! If you want to show that this is true for any element in the set, you should not have used the word choose in the proof's beginning, or, at least, you should say at the end that since our intermediate steps do not use the special assumption of $a,$ it suits for all elements. But in this way, this becomes slightly complex than Method 1, why not just use Method 1?" The reason that Method 2 is valid is somewhat of a meta-level, not indicate by the language itself.

ryang
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Eric
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    It's not because you just happened to luckily choose the right element in approach 2 - you use properties that they all have – The Chaz 2.0 Oct 06 '16 at 17:53
  • The English in your way (2) is close to unintelligible (even after I've guessed that "U.G." stands for "universal generalisation"). If you present way (2) in good idiomatic mathematical English (writing "let us choose some $a$ in $D$, rather than "Arbitrary choose $a$ in $D$ and so on), there is no substantive difference from way (1). – Rob Arthan Oct 07 '16 at 22:16
  • @RobArthan Yes!! So do you also think that begin a proof with "Arbitrary choose $a$ in $D$ and so on " is not good? – Eric Oct 08 '16 at 05:54
  • 'Arbitrary' and 'any' mean the same in the above context. If you don't like the word "choose" because you feel that it conflicts with arbitrariness, and you want an alternative to the word "let", then my answer here suggests writing "Consider any $a\in D$" or "Consider an arbitrary $a\in D$". – ryang Feb 25 '23 at 20:08

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The two wordings mean exactly the same and it's up to personal preferences which variant phrasing to use.

As for your objection to the second phrasing, "arbitrarily" does not mean "randomly". If you want, you can imagine that the author is asking you to choose an element arbitrarily, and in the proof he then shows that no matter what you arbitrarily chose, it will have the desired property.

hmakholm left over Monica
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  • A perfect example of this is the classic "Let $\epsilon > 0$" deal. No matter what $\epsilon$ is the condition will still hold provided the proper $\delta$ or $N$ or whatever is chosen. – Sean Roberson Oct 06 '16 at 17:54
  • but in that moment, arbitratily choose appears to mean(at least in fact we) randomly choose a particular one. -- The choice is arbitrary, but after picking one, the following all argument is do for that particular one. To connect it to mean "for all element it also true", we seems to have more words to explicitly state. – Eric Oct 06 '16 at 18:09
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    @Eric: "Arbitrary" does not mean "random". Where your insistence that the choice has to be random comes from I don't know, but it is not from the text. – hmakholm left over Monica Oct 06 '16 at 18:11
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    Henning, when trying to understand "arbitrary" the colloquial use of "random" comes to mind; @Eric: Do note that "random" has a mathematical meaning, and this is not the correct context to be using it. Which is probably why Henning is a bit confused about your usage of the word. – Asaf Karagila Oct 07 '16 at 07:41
  • @AsafKaragila I see. I think "arbitrary choose", is to choose, though any, but one, element in the set, and start your all argument on that particular element. Yet, at the end of the proof, one can say "because any element in the set also fit our argument, so this proof is valid for all elements. Q.E.D" However, that the word "choose" have become a little bit unnecessary, as my post said. It can be more easy to state a proof begin with "let $a$ be any element in the set", the tone is more natural and causal about we don't really care what we had chose; whereas the word "choose" is not. – Eric Oct 07 '16 at 13:16
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    @Eric: Again, there is a lot of personal style here. When you've written "Let $a$ be an arbitrary element" for the thousandth time, you'll start looking for alternatives to spice up your proofs. I've developed (and healed from) allergies to the word "therefore" several times in my years as a grad student. – Asaf Karagila Oct 07 '16 at 13:19