I'm having some difficulty visualising the difference between the limit supremum and supremum (and for limit infimum/infimum) for bounded sequences. Would it be possible for some to provide a brief explanation and maybe some examples?

  • $\begingroup$ Sup you can calculate to sets, not to sequences $\endgroup$ – Martín Vacas Vignolo Apr 9 '16 at 1:10
  • 1
    $\begingroup$ @vvnitram can you elaborate a bit? $\endgroup$ – Inazuma Apr 9 '16 at 1:11

Let $(x_n)$ be a bounded sequence. The supremum is the the least upper bound of the sequence as a set. For the limit-supremum, or $\limsup$, is the limit of the sequence $(a_n)$, where $a_n=\sup_{k\geq n}x_k$. In plainer language, the limit supremum measures how the supremum of the sequence behaves as we start removing terms from the sequence, starting from the beginning. The nice thing about $\limsup$ is that it always exists, even if the sequence doesn't converge at all.

Sometimes the $\limsup$ is equal to the $\sup$. Take for example the sequence $x_n=1-1/n$, where $\sup\{x_1,x_2,x_3,...\}=1$. If we take the $\sup$ of the subsequence $\{x_n,x_{n+1},x_{n+2},...\}$, we still get $1$ for each $n$, so $\limsup\{x_n\}=1$.

Other times, $\limsup$ is less than $\sup$. If $x_n=1/n$, $\sup\{x_1,x_2,x_3,...\}=x_1=1$. If we remove the first term, $\sup\{x_2,x_3,x_4,...\}=x_2=1/2$. Similarly, the more terms we remove, the lower the $\sup$ is. As we remove more and more terms, the $\limsup\{x_n\}=0$. The $\limsup$ is never greater than $\sup$. See if you can prove this to yourself.


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