I'm a bit confused with the general concept of convergence of a sequence of sets.

I'm well aware that the limit of a sequence $\{C^{\nu}\}$ exists iff $$\liminf_{\nu \rightarrow \infty} C^{\nu} = \limsup_{\nu \rightarrow \infty} C^{\nu}$$ where lim inf (resp. lim sup) is the set of points that appear in the limit all but finitely many times (resp. infinitely many times).

However, intuitively, the limit point can appear only once, i.e., for $\nu \rightarrow \infty$. Isn't this in contrast with the concepts of lim inf and lim sup (defined as above)?

For instance, let $C^{\nu} \triangleq [0,1-1/\nu]$: the sequence $\{C^{\nu}\}$ should (intuitively) converge to $C \rightarrow [0,1]$. However, I think the point $\{1\}$ is included in $C$ only for $\nu=\infty$ and, therefore, it appears only once.

What am I missing?

  • $\begingroup$ The sequence $[0,1-1/\nu]$ converges to $[0,1)$. $\endgroup$ – Cm7F7Bb Mar 5 '14 at 14:20
  • $\begingroup$ $\nu$ does not take 'value' $\infty$. $\endgroup$ – drhab Mar 5 '14 at 14:29
  • $\begingroup$ @V.C.Are you absolutely sure? I have been told otherwise here by Jeff Snider math.stackexchange.com/questions/699384/… (see comments) $\endgroup$ – TheDon Mar 5 '14 at 14:32
  • $\begingroup$ @ItaAtz Yes, I'm sure. Since the sequence $[0,1-1/\nu]$ is monotone non-decreasing, we have that $\lim_{\nu\to\infty}[0,1-1/\nu]=\cup_{\nu=1}^\infty[0,1-1/\nu]=[0,1)$. $\endgroup$ – Cm7F7Bb Mar 5 '14 at 14:43
  • $\begingroup$ @V.C. Can I have a reference to this statement? As a general comment, can you suggest me a good reference (book, or whatever), other than "Variational Analysis" by Rockafellar and Wets (and possibly simpler) to study these concepts of set convergence theory? Thanks a lot $\endgroup$ – TheDon Mar 5 '14 at 14:51

There are two competing definitions of $\liminf$ and $\limsup$. The one referenced by @V.C. in the comments is this: $$\liminf_{\nu\to\infty} C_\nu := \bigcap_{n\in\mathbb{N}}\bigcup_{\nu\ge n}C_\nu$$ which we can rewrite in the notation of Rock & Wets as $$\liminf_{\nu\to\infty} C_\nu := \bigcap_{N\in\mathcal{N}_\infty^\#}\bigcup_{\nu\in N}C_\nu.$$ The definition from Rock & Wets, and other sources such as Hocking & Young, adds a closure: $$\liminf_{\nu\to\infty} C_\nu := \bigcap_{N\in\mathcal{N}_\infty^\#} \text{cl}\bigcup_{\nu\in N}C_\nu.$$ (For $\limsup$ simply transpose the union and intersection operators.)

For what it's worth, Wikipedia uses the latter definition, written equivalently as:

lim inf Xn, which is also called the inner limit, consists of those elements which are limits of points in Xn for all but finitely many n (i.e., cofinitely many n). That is, x ∈ lim inf Xn if and only if there exists a sequence of points {xk} such that xk ∈ Xk and xk → x as k → ∞.

Note that in a discrete space the two definitions coincide. $\mathbb{R}^n$ with the usual topology is not a discrete space.

  • $\begingroup$ Thanks @Jeff Snider, I've been fighting with this dilemma for days. Perhaps you'd like to take a look at this other question of mine: math.stackexchange.com/questions/698966/… It's the problem for which (very recently) I had to start studying set convergence theory. $\endgroup$ – TheDon Mar 5 '14 at 19:16
  • $\begingroup$ Actually I've just noticed (from Rock & Wets Exercize 4.2(b)) that, to obtain the equivalent formula for the limsup, it's not just swapping the union and intersection operators but rather considering the whole set or the its "tail". $\endgroup$ – TheDon Mar 6 '14 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.