limit of sequence of sets

Recall that lim sup and lim inf of a sequenceMathworldPlanetmathPlanetmath of sets {Ai} denote the and the of {Ai}, respectively. Please click here ( to see the definitions and here ( to see the specialized definitions when they are applied to the real numbers.

Theorem. Let {Ai} be a sequence of sets with i+={1,2,}. Then

  1. 1.

    for I ranging over all infinite subsets of +,

    lim supAi=IiIAi,
  2. 2.

    for I ranging over all subsets of + with finite compliment,

    lim infAi=IiIAi,
  3. 3.

    lim infAilim supAi.


  1. 1.

    We need to show, for I ranging over all infinite subsets of +,

    IiIAi=n=1i=nAk. (1)

    Let x be an element of the LHS, the left hand side of Equation (1). Then xiIAi for some infinite subset I+. Certainly, xi=1Ai. Now, suppose xi=kAi. Since I is infiniteMathworldPlanetmath, we can find an lI such that l>k. Being a member of I, we have that xAli=k+1Ai. By inductionMathworldPlanetmath, we have xi=nAi for all n+. Thus x is an element of the RHS. This proves one side of the inclusion () in (1).

    To show the other inclusion, let x be an element of the RHS. So xi=nAi for all n+ In i=1Ai, pick the least element n0 such that xAn0. Next, in i=n0+1Ai, pick the least n1 such that xAn1. Then the set I={n0,n1,} fulfills the requirement xiIAi, showing the other inclusion ().

  2. 2.

    Here we have to show, for I ranging over all subsets of + with +-I finite,

    IiIAi=n=1i=nAk. (2)

    Suppose first that x is an element of the LHS so that xiIAi for some I with +-I finite. Let n0 be a upper boundMathworldPlanetmath of the finite setMathworldPlanetmath +-I such that for any n+-I, n<n0. This means that any mn0, we have mI. Therefore, xi=n0Ai and x is an element of the RHS.

    Next, suppose x is an element of the RHS so that xk=nAk for some n. Then the set I={n0,n0+1,} is a subset of + with finite complement that does the job for the LHS.

  3. 3.

    The set of all subsets (of +) with finite complement is a subset of the set of all infinite subsets. The third assertion is now clear from the previous two propositionsPlanetmathPlanetmath. QED

Corollary. If {Ai} is a decreasing sequence of sets, then

lim infAi=lim supAi=limAi=Ai.

Similarly, if {Ai} is an increasing sequence of sets, then

lim infAi=lim supAi=limAi=Ai.

Proof. We shall only show the case when we have a descending chain of sets, since the other case is completely analogous. Let A1A2 be a descending chain of sets. Set A=i=1Ai. We shall show that

lim supAi=lim infAi=limAi=A.

First, by the definition of of a sequence of sets:

lim supAi=n=1i=nAk=n=1An=A.

Now, by Assertion 3 of the above Theorem, lim infAilim supAi=A, so we only need to show that Alim infAi. But this is immediate from the definition of A, being the intersectionMathworldPlanetmathPlanetmath of all Ai with subscripts i taking on all values of +. Its complement is the empty setMathworldPlanetmath, clearly finite. Having shown both the existence and equality of the and of the Ai’s, we conclude that the limit of Ai’s exist as well and it is equal to A. QED

Title limit of sequence of sets
Canonical name LimitOfSequenceOfSets
Date of creation 2013-03-22 15:00:34
Last modified on 2013-03-22 15:00:34
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 8
Author CWoo (3771)
Entry type Theorem
Classification msc 03E20
Classification msc 28A05
Classification msc 60A99
Related topic LimitSuperior