While working on a problem, I was wondering about the following: Is it possible to write any infinite set as union of finite sets or not ?
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2$\begingroup$ Congrats @Stefan: for beating Asaf to the punch for changing a (set-theory) tag to (elementary-set-theory) :P $\endgroup$– Eric StuckyNov 4, 2016 at 3:19
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5$\begingroup$ @EricStucky Well, that's one lifetime achievement taken care of :P $\endgroup$– Stefan MeskenNov 4, 2016 at 4:17
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$\begingroup$ Hahaha @EricStucky/Stefan is this a thing now? :) $\endgroup$– user541686Nov 4, 2016 at 4:39
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2$\begingroup$ If we could, why bother with having an axiom to define an infinite set? $\endgroup$– JDługoszNov 4, 2016 at 7:57
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$\begingroup$ @JDlugosz: If there aren't any infinite sets, then it's automatically true that every infinite set can be written as a union of finite sets. $\endgroup$– user14972Nov 4, 2016 at 17:46
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1 Answer
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Sure, it's possible. Let $X$ be any set. Then $$ X = \bigcup_{x \in X} \{x \}. $$ So any set is a union of singletons (except maybe for the empty set - dependent on how you'd interpret the above formula in this case).
answered Nov 4, 2016 at 1:16
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1$\begingroup$ @Sara Yup! Nothing different there. "$\bigcup_{x\in X}A_x$" always means "The set of things which are in some $A_x$ (with $x\in X$)", so "$\bigcup_{x\in X}\{x\}$" always means "$X$." $\endgroup$ Nov 4, 2016 at 2:04
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4$\begingroup$ @Stefan The empty union is empty, so that still works. ($\bigcup_{x\in X}A_x=\{y: \exists x\in X(y\in A_x)\}$, so if $X$ is empty $\bigcup_{x\in X}\{x\}$ is just $\emptyset$.) $\endgroup$ Nov 4, 2016 at 2:05
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1$\begingroup$ @Noah Yes, I'm aware ;) But I also know that some mathematicians feel uneasy about 'the empty union' much like they feel uneasy about 'the empty function'. $\endgroup$ Nov 4, 2016 at 2:17
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9$\begingroup$ @Stefan: Some mathematicians feel uneasy by anything empty, it is called fear of the void. This is in large part because teaching systematically avoids talking about them for fear of making the audience uneasy. Think of empty matrices in linear algebra, where (sub)spaces of dimension $0$ (which aren't even empty) are of eminent importance. In the current case there is no reason to shy away from the empty union: the union of elements of any set exists. What is problematic is the empty intersection, and it is good to be aware of that. But this difficulty is not contagious. $\endgroup$ Nov 4, 2016 at 5:45
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1$\begingroup$ @chi: But the question only asked if something was true for any infinite set, and for that you don't need to prove that infinite sets exist; in ZFC minus Infinity, you have models without infinite sets, but in those models the claim is vacuously true, and models with infinite sets, which are then the same you get with full ZFC. So you do not need the axiom of infinity. Indeed, you just need three axioms: Extensionality (without which you'd not be able to tell whether the union is the same set), Union (obviously) and Replacement (to get from $S$ to $\{\{x\}:x\in S\}$. $\endgroup$– celtschkNov 4, 2016 at 16:12