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In many proofs I see that some variable is "fixed" and/or "arbitrary". Sometimes I see only one of them and I miss a clear guideline for it. Could somebody point me to a reliable source (best a well-known standard book) which explains, when and how to use both in proofs?

EDIT: A little add-on to the question: Take a usual induction on natural numbers and assume that you are teaching it to students in their first semester. How do you explain "fixed" and "arbitrary" in this scenario?

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    $\begingroup$ As Quiaochu Yuan points out, one place you will see this is beginning (and sometimes not so beginning) analysis. It is an attempt partly to defang the logical complexity of $\forall\epsilon\exists \delta\forall x$. At the earliest level, the student may even be invited to think of the "fixed" $\epsilon$ as being, say, $1/1000$. $\endgroup$ – André Nicolas Jun 21 '11 at 16:14
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    $\begingroup$ bln-tom: To address your second question, I did that during the fall semester. The answer I can give you is simply "carefully.", it is also important to explain how these words are not magical. Writing them will not make a bad proof be correct (a common freshman mistake). $\endgroup$ – Asaf Karagila Jun 22 '11 at 6:36
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Both "arbitrary" and "fixed" are just shorthand for a universal quantifier. When I say something like "fix $\epsilon > 0$" it means I am about to prove a statement that is true for all $\epsilon > 0$ (and thereby prove that some function is continuous, for example) but I don't want to actually write out "for all $\epsilon > 0$" in front of every sentence I'm about to write. That's really all there is to it.

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    $\begingroup$ I agree that this is confusing though; at first glance "arbitrary" and "fixed" sound like they mean opposite things, and then some people say "fix an arbitrary..." $\endgroup$ – Qiaochu Yuan Jun 21 '11 at 16:00
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When you encounter the term "arbitrary", it usually just means that a given statement is specified for any element from a given set of elments. For example, if I say, let $x$ be an arbitrary element of the interval $[0, 1]$ I just mean that $x$ can take on any value within that interval.

The term fixed connotes a similar but more specific meaning. If I say let $x$ be a fixed element of the interval $[0, 1]$ I mean that, firstly, $x$ is an arbitrary element of $[0,1]$ but it's value is unchanging througout its usage.

The difference in meaning is subtle but sometimes important. If you start out with a "fixed" element you can't, for example, choose its value at some later point in the argument. On the other hand, if I prove that something holds for an arbitrary element of a given class, then it will hold for any particular element of that class which can often be very useful.

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    $\begingroup$ Perhaps an example would be helpful in your last paragraph: Sometimes in an $\epsilon$-$\delta$ proof one will start off by proving certain propositions "for arbitrary $\epsilon > 0$", and later choose a value of $\epsilon$ that makes the proof go through. It would be very jarring to do this if one had started with "fixed $\epsilon$". $\endgroup$ – Rahul Jun 21 '11 at 16:21
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    $\begingroup$ @Rahul Nice Example $\endgroup$ – ItsNotObvious Jun 21 '11 at 16:47
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    $\begingroup$ Are you confusing arbitrary and fixed with free and bound variables? $\endgroup$ – Isaac Kleinman Jul 17 '13 at 23:44
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Rather then arbitrary and fixed, lets look at something very similar where the difference is more pronounced. There is a very important distinction between uniform and fixed. A perfect example is convergence of functions.

Let $f_n$ be a sequence of functions of a set $X$. For each point $x\in X$, let $f$ be defined by $f(x)=\lim_{n\rightarrow \infty} f_n(x)$. That is, $f$ is the pointwise limit, and it is the limit for fixed $x$.

However, uniform convergence is much stronger, and requires that the limit is small for all $x$ simultaneously.

Very often things come up which are easy to prove for fixed $x$, but very difficult when we need uniformity in $x$.

Hope that helps,

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    $\begingroup$ Although your answer is correct it adds on the confusion, especially since all the answers given so far (yours, Qiaochu's and 3Sphere's) are all valid. You should add some further clarification on the matter. $\endgroup$ – Asaf Karagila Jun 21 '11 at 16:20

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