I'm taking real analysis and I still find this concept hard to understand:

If S is any bounded nonempty set of real numbers, $\sup(S)$ and $\inf(S)$ exist.

However, consider a bounded decreasing sequence $\{a_n\}$. Take $a_n= \frac 1 n$). Consider $\{b_n\}$ given by $b_n= \sup \{a_k: k \ge n \}$. We could prove $b_n$ was monotonic and bounded, thus converges. We wrote $\limsup a_n =\lim b_n$. In our case, $\lim b_n = \lim \frac 1 n = 0$.

However, $b_n$ is the superior of $\{a_n\}$, as the least upper bound of $a_n$, thus $b_n$ must be bigger or equal to $a_n$, which contradicts the result. 

I searched online for limit superior but couldn't understand the notation well.

Could you help me to explain:

  1. What is the real definition of limit superior in $\mathbb{R}$?
  2. What went wrong, and the correct answer to the question?
  • $\begingroup$ $\lim\!\sup$ is not the same as $\sup$. $\endgroup$ Feb 15 '17 at 1:27
  • $\begingroup$ In your Definition, you have $a_n = b_n$ as $a_k$ is decreasing. $\endgroup$
    – user251257
    Feb 15 '17 at 1:29

The real definition of $\limsup$ is the one you wrote down: $$ \limsup_{n\to\infty} a_n = \lim_{n\to\infty} b_n $$ where $b_n=\sup(\{a_k|k\ge n\})$

Why does $b_n$ being the least upper bound of the tail cause a contradiction? It's certainly true that $b_n\ge a_n$ but I don't see how that's a problem.

It's totally possible and commonplace for the limit (and limsup) of a sequence to be greater than any element of the sequence. For instance $\lim_n (1-1/n)= \limsup_n(1-1/n)=1,$ and we have $1>a_n$ for all $n$.

Also, addressing the title of your question, I think you may be confused. The limsup of a sequence always exists (in $[-\infty,\infty]$). No need for boundedness or monotonicity. Where monotonicity comes in is in the proof of this. $b_n,$ as defined above, is a monotonically decreasing sequence in $[-\infty,\infty]$, and thus, by a theorem, always has a limit in $[-\infty,\infty].$ If the sequence $a_n$ is bounded (i.e. there is a $B\in \mathbb R$ such that $|a_n|<B$ for all $n$), then $b_n$ is a monotonically decreasing bounded sequence in $\mathbb R$ and thus by another version of the same theorem, the limit exists in $\mathbb R.$

So the only way your title makes sense to me is if you're trying to prove that the limsup exists in $\mathbb R$ (as opposed to $[-\infty,\infty]$) and in that case you only need boundedness of $a_n,$ not monotonicity.


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