Given a bounded sequence of real numbers

I want to show $x_n \in [\liminf(x_n) - K, \limsup(x_n) + K]$ for all $n\geq N,$ where $N$ is some natural number, and for all $K>0.$

Attempt at a solution:

Since the $\limsup$ and $\liminf$ are accumulation points, there are subsequences that converge to each on respectively. Call them $x_{nl}$ and $x_{nm}$

Thus given $K > 0$, there exists $N_1$ such that $|x_{nl} - \text{lim inf}| < K$ for all $n_l \geq N_1,$ and similarly there exists an $N_2$ for which $|x_{nm} - \text{lim sup}| < K$ for all $n_m \geq N_2$

Now consider $M = \max \{N_1,N_2\}$. Then all terms of those $2$ subsequences that are after this number $M$ must fall in $\left[\liminf(x_n) - K, \limsup(x_n) + K\right].$

Now consider the terms of $x_n$ for $n \geq M$ that do not fall into one of these $2$ subsequences, if there are finitely many, then we can simply consider $M^* = \text{max} \{N_1,N_{2,j} \}, $ where $j$ is the largest index for which $x_j$ does not belong to one of the two subsequences.

Now we must consider the case where infinitely many terms don't fall into one of these two subsequences and this is where I am stuck.
I considered taking the subsequence composed of all such terms and trying to show it itself has a subsequence which is convergent and thus has an accumulation point which falls between $\liminf$ and $\limsup$ but I do not know how that will lead me to be able to conclude that in general all terms must fall between $\liminf - K$ and $\limsup + K.$

Any help to get past this step would be very appreciated.

  • 1
    $\begingroup$ Use the commands $\limsup$, similarly for max and lim inf, to make them look nicer. Also, use $\leq$ for less than or equal to. $\endgroup$ Oct 27, 2018 at 17:20
  • $\begingroup$ Thanks, I was thinking that there must be a better way to do it! I'm new to MathJax and I appreciate all tips. $\endgroup$
    – mmmmo
    Oct 27, 2018 at 17:43

1 Answer 1


I think I may have found the theorem I was searching for. Let $T_m$ be the upper m-tail of $(x_n)$ and define $V_m$ = $sup(T_m)$. Then $lim(V_m)$ = $\limsup(xn)$.

With this theorem we can now see there must be some $m_1$ for which $|V_m - \lim sup(xn)| < \epsilon$ is true for all $m \geq m_1$ and thus $|V_m - \limsup(x_n)| < \epsilon$ and as a result $x_n - \limsup(x_n) < \epsilon $ for all n $\geq$ $m_1$

Similarly, we can do the same thing for $\lim inf$ to find that there is some $m_2$ for which $\liminf - x_n < \epsilon$ for all $m \geq m_2$

If we take M to be the max of $m_1$ and $m_2$ then we have found the result we wanted


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.