How is arbitrary different from finite in mathematics? An arbitrary union of subsets and a finite union of subsets look the same to me! Is an infinite union of subsets arbitrary?

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    $\begingroup$ Arbitrary unions include both finite unions and infinite unions. $\endgroup$ – GEdgar Feb 19 '13 at 22:36
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    $\begingroup$ It's probably helpful to note that usually countable is stuck in between. $\endgroup$ – Sean Ballentine Feb 19 '13 at 22:39

The meaning is totally different. The statement "for arbitrary $x$" means "for all $x$", whereas "finite" is a term that can be applied to a set to indicate that its cardinality (size) is a natural number.

In particular, asking "is $A$ arbitrary" only makes sense in certain contexts, where you may be asking whether something is being proved for all $A$, or only for a specific $A$.

Sometimes the phrase "arbitrary union" is used as shorthand for "union of a collection of arbitrary cardinality" as opposed to "union of a finite collection" or as opposed to "union of a pair" (binary union.)


The word "arbitrary" does not specify a property of the thing it applies to. It is merely a rhetorical way to call attention to the fact that what you say is not restricted to things with any particular property.

It's still left to the reader to figure out from context what the thing you're calling attention to not saying is, but that is okay, because the word "arbitrary" has no formal effect.

So saying that "such-and-such is true for arbitrary unions" formally means neither more nor less than saying that "such-and-such is true for unions", period.

When we say "let $A$ be an arbitrary widget", we're not formally saying anything different from "let $A$ be a widget" -- just making sure that the reader notices that the only thing we know about $A$ is its widgetness. We can follow up with "remember that $A$ was arbitrary ..." which again just mean "remember that we didn't assume anything in particular about $A$".


Just to add to what Aloizio and Trevor have said -

Suppose I say "let $U$ be an arbitrary subset of the real line." Then $U$ can be anything. Some choices:

  • $U=$ all reals

  • $U=$ all rationals

  • $U=$ all irrationals

  • $U=$ all numbers greater than $\sqrt{2}$

  • $U=$ all even integers

  • or perhaps $U=$ just the set $\{1,2,3,4,19\}$

  • or anything else you can dream up.

Suppose I say "let $U$ be a finite subset of the real line." Now that means $U$ must only have a finite number of elements. Of the above choices, the only one that still qualifies is $U=\{1,2,3,4,19\}$.


Arbitrary means you can have any indexed union of sets.

Finite means you can have only a finite union, i.e., you can only have an union indexed by a set of same cardinality of $I_n$, where $I_n:=\{1,...,n\}$

For example:

Define $A_{\alpha}:=[\frac{1}{\alpha}, 1]$ for every $\alpha \in [1, \infty)$

You can make $\displaystyle \bigcup _{\alpha \in[1,\infty]} A_{\alpha}$, and it will be a good exercise for you to show that this is $(0,1]$

  • $\begingroup$ Just an observation... my answer was heavily induced by the use of "arbitrary" in the context you mentioned especifically. My colleagues' answers are far more general and will answer your question more thoroughly. $\endgroup$ – Aloizio Macedo Feb 20 '13 at 0:31
  • $\begingroup$ Are you sure that's a union (the exercise)? Because $A_1=\left[1,1\right]$. $\endgroup$ – saadtaame Feb 20 '13 at 12:38
  • $\begingroup$ I think that that's an intersection. Here is my attempt at proof: $$\bigcap_{\alpha=1}^{\infty}A_\alpha=\lim_{\alpha \to \infty}A_\alpha=\left (0,1\right ]$$ $\endgroup$ – saadtaame Feb 20 '13 at 12:45
  • $\begingroup$ It is a union. Be careful with some things and "sweeping generalizations". First, why is the first equality you use true? (Important observation: that notation you used in the intersection is usually used for countable intersection, which is not the case). Second, what even means that $\displaystyle \lim _{\alpha \rightarrow \infty} A_{\alpha}$? Now back, to prove my statement, you must prove that: (1) $\displaystyle (0,1] \subset \bigcup _{\alpha \in [1,\infty)} A_{\alpha}$ and (2) $\displaystyle \bigcup _{\alpha \in [1,\infty)} A_{\alpha} (0,1] \subset (0,1]$ $\endgroup$ – Aloizio Macedo Feb 22 '13 at 13:40
  • $\begingroup$ I'll do (1) as an example: $\displaystyle x \in (0,1] \Rightarrow 0<x\leq 1 \Rightarrow \frac{x}{2}<x\leq 1 \Rightarrow \frac{2}{x}>1 ~~~~~~~~~~$ $\displaystyle \frac{2}{x}>1 \Rightarrow \frac{2}{x} \in [1,\infty)~~~~~~~~$ $\displaystyle A_{\frac{2}{x}}=[\frac{x}{2}, 1]$ Therefore, $\displaystyle x \in A_{\frac{2}{x}}$, and then we have that $\displaystyle x \in \bigcup _{\alpha \in [1,\infty)} A_{\alpha}$ $\endgroup$ – Aloizio Macedo Feb 22 '13 at 13:49

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