Given any uncountable set S, would I need to use transfinite induction to prove if I remove single elements recursively, I will be left with the empty set?

It seems like this can be thought of as an arbitrary intersection problem, so just a logical consequence. Could I do with something weaker than transfinite induction?

  • $\begingroup$ +1 A good question, trying to understand some difficult definitions. $\endgroup$ May 1, 2013 at 3:25

2 Answers 2


Yes and no. If you remove single elements recursively, you will eventually remove everything, but you need a wellordering to be able to do that in the first place. To remove single elements recursively, you have to do it in some order; it is a transfinite recursive operation. To show that it eventually removes everything, you would probably still have to use transfinite induction in a more or less veiled way, but you don't need any more “choice” to do that.

On the other hand, if you just remove every single element without caring about the order, you don't need anything like that. $\bigcap_{s\in S} S\setminus \{s\}=\emptyset$, period.

  • $\begingroup$ I changed the quotes on choice because the opening ones were below the line. I couldn't get nice opening and closing ones. If you prefer the original, feel free to roll back. $\endgroup$ May 1, 2013 at 3:36
  • $\begingroup$ @RossMillikan: In Polish, we write opening quotes at the bottom, so I wrote it like this out of habit. I now replaced it according to the usual English convention (with opening quote at the top). :) $\endgroup$
    – tomasz
    May 1, 2013 at 8:05
  • $\begingroup$ I am a little confused by your answer. In your second sentence, without mentioning order, you say just to removing everything recursively I need a well ordering. However, in your second paragraph, you say if I don't care about order, then I can do this without using transfinite induction. Would I be correct in interpreting that if I successively remove elements at random, I do not need a well-ordering, but if if I remove elements in accordance with some pattern, then I must have the set well-ordered first? If this is true, this definitely makes sense to me. $\endgroup$ May 1, 2013 at 21:38
  • 1
    $\begingroup$ @user75294: You need some wellorder to make sense of "removing elements successively". In a way, in the second paragraph, elements are not removed successively, they are removed all at once. $\endgroup$
    – tomasz
    May 3, 2013 at 13:32
  • $\begingroup$ Good. This makes sense. Thank you! $\endgroup$ May 3, 2013 at 15:20

You can't remove a countably infinite number of elements one element at a time. Consider the following family of sets: $A_0 = N$ and $A_{i+1} = A_i - \{i\}$. The empty set can't be a member of this family. Assume $A_{z+1} = \{\}$. Then $A_z = \{z\}$ and $z$ is the largest natural number. I don't see how assuming $A_0$ is uncountable would change anything.

Edit in response to comments.

Tomasz is correct this can be proven using arbitrary intersection. Show there is a collection of sets such that every element of S is missing from some set in the collection. The intersection of this collection will be the empty set. What you can not do is "remove single elements recursively".

The intersection of a collection of sets is a single operation. This is why $\cap_{k<\omega} A_k = \{\}$. Recursively taking the intersection of two sets at a time will never give us the empty set.

$(A_0 \cap A_1) = A_1$

$(A_0 \cap A_1) \cap A_2) = (A_1 \cap A_2) = A_2$


$\cap_{k \leq n} A_k = A_n$

Induction does not help either. We can easily prove $\forall A_x \exists y(y \in A_x))$ using induction. The basic problem is the empty set is a finite set. We can never get a finite set by removing one element from an infinite set.

  • $\begingroup$ You have in fact removed a countably infinite number of elements one at a time: $\bigcap_{k<\omega}A_k=\varnothing$, and the recursive construction cannot continue. Had you removed $2k$ at step $k$, you’d have removed all of the even non-negative integers, leaving the odd ones, and the construction could have been continued for at least another $\omega$ stages, depending on how you did it. $\endgroup$ May 1, 2013 at 14:35
  • $\begingroup$ We commonly deal with arbitrary (so, perhaps, uncountably many) intersections, so it seems perfectly valid to concluded if we perform such an operation on a countable set, we will be left with the empty set. Consider, for example, successive intersections of the increasing sequence [-k,k] for each k as k tends to infinity. We end with (-inf,inf). The reals subtract this is the empty set. $\endgroup$ May 1, 2013 at 21:43
  • $\begingroup$ Thank you Russel for your edits. This has been extremely enlightening for me. If one interprets $\bigcap_{\forall m\ge n} [n,\infty)=\emptyset$ as single elements being removed successively, then there exists an infinite set such that removing one element from it gives a finite set, which is impossible. Of course, what I meant was what tomasz said, but formally speaking, you are correct. I must admit, I disagreed at first, and was going to cite von Neumann's construction of the ordinals as an example of a single operation being repeated "uncountably" many times, but his (continued -> $\endgroup$ May 3, 2013 at 0:54
  • $\begingroup$ construction requires transfinite induction, which involves limit ordinals, and these cannot be "made" one element at a time. This is precisely the difference between transfinite induction and regular induction. Is all this correct? $\endgroup$ May 3, 2013 at 0:56
  • $\begingroup$ Thanks for the great question. First order induction is <a href="math.stackexchange.com/questions/328984/…> with the axioms of $ZF$. Transfinite induction needs limit ordinals. This "hides" the use of intersection or union inside the definition of the limit ordinal. $\endgroup$ May 4, 2013 at 7:11

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