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Generally, when one is going to prove a result regarding a set of elements, they begin their proof with those first few pleasing words: "Suppose...is an arbitrary element in..."

My question is, why does considering an arbitrary element in a proof imply that the proven result applies to every element in the set? Although I think I have an vague about this, I am still interested in seeing what others have to say in relation to this idea. Perhaps, if possible, answer the question in the manner that you would if a student of yours posed this question to you.

Thank you!

Steven Stadnicki
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Mack
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  • I've changed the title so that it doesn't need to be fleshed out with the 'long-enough' text; hopefully I've preserved the core meaning of the question, but if this doesn't suit you, feel free to change it back. – Steven Stadnicki Jun 13 '14 at 15:44
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    It's interesting to note that in Euclid's Elements, at least in the Heath translation, whenever Euclid has a "Let $x$ be any $X$...", he felt it necessary to add "And similarly we can prove the same for any other $X$." after his argument. – Jack M Jun 13 '14 at 21:56

4 Answers4

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You could instead say "let $x$ be any element of our set" that is the same as saying "let $x$ be an arbitrary element of our set". This just means we are not assuming anything about $x$ other than it is in our set. So, anything we prove about $x$ holds for any element in our set.

John Machacek
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    I particularly like when you said "This just means we are not assuming anything about x..." That's an interesting idea. – Mack Jun 13 '14 at 15:49
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    That is, we are not assuming anything beyond that which we explicitly state. – Jørgen Fogh Jun 14 '14 at 11:15
  • I know this is very late, but could I ask what x is precisely? Is it a 'mathematical object', in the same sense as the integer 2? It seems like x can be substituted with elements from a set rather than being an element of a set itself? – masiewpao Oct 24 '19 at 15:52
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Recall that a statement of the form $\forall x\varphi(x)$ is true in a structure $M$ if and only if for every $m\in M$ it holds that $\varphi(m)$ is true.

In the proof we usually show that there is a schema which can prove for each $m\in M$ that $\varphi$ holds. And so we can show that $\forall x\varphi(x)$ holds in $M$.

Asaf Karagila
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This will be clarified if you study formal logic. For a nice introduction see the entry on Classical Logic in the Stanford Encyclopedia of Philosophy. The article addresses your question explicitly in a comment following the $\,\forall$-introduction rule, $ $ namely

This rule (∀I) corresponds to a common inference in mathematics. Suppose that a mathematician says “let n be a natural number” and goes on to show that n has a certain property P, without assuming anything about n (except that it is a natural number). She then reminds the reader that n is “arbitrary”, and concludes that P holds for all natural numbers. The condition that the variable v not occur in any premise is what guarantees that it is indeed “arbitrary”. It could be any object, and so anything we conclude about it holds for all objects.

See also the following section for the semantics of universal quantificaton, and the final section on Meta Theory for relationships between deductive systems and their models (e.g. soundness).

Bill Dubuque
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  • I'm not sure this really clarifies it. You're just giving it a name. – Jack M Jun 13 '14 at 21:05
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    @JackM It's not just a name. Rather, it is a rule in a deductive system. Such rules serve to specify the meaning of universal quantification in the deductive system. These rules serve to axiomatize the notion of universal quantification in first-order logical systems, analogous to how the ring axioms serve to axiomatize the operations of addition and multiplication in "number systems". – Bill Dubuque Jun 13 '14 at 21:47
  • "Rule in a deductive system" is the "name" I was referring to. – Jack M Jun 13 '14 at 21:54
  • @JackM The point was to show how the intuition is formally (rigorously) captured in proof theory. – Bill Dubuque Jun 13 '14 at 22:01
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    @JackM The block quote, particularly the part from "She then reminds" to the end exactly answers the question. – David Richerby Jun 14 '14 at 16:40
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Let's say we have an argument. You say "for all x in the set S, A (x) is true". I say "I don't believe it. I'm sure there is an x in the S where A (x) is not true". You say "well, tell me an x and I'll show you A (x) is true". I say "I can't think of any x right now, so just assume any x". You say "Ok then, we will assume that x is any element of the set S, and I'll prove that A (x) is true. And since we don't make any assumption about that x, the same proof will work for any x in the set, so S (x) is true for all x".

gnasher729
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