How can we prove that $\liminf x_n = -\limsup (-x_n)$?

The definitions we are using are
$\limsup x_n = \lim\limits_{n\to\infty} \sup\{x_k; k\ge n\}$
$\liminf x_n = \lim\limits_{n\to\infty} \inf\{x_k; k\ge n\}$

  • 3
    $\begingroup$ What definition of $\liminf$ and $\limsup$ do you take? What did you try? $\endgroup$
    – Julien
    Mar 18, 2013 at 20:35
  • 1
    $\begingroup$ first you need to prove that Inf (-A)=-Sup (A) $\endgroup$
    – cswinson
    Mar 18, 2013 at 20:37
  • $\begingroup$ lim Sup xn= lim (Sup {xn : n>N}) $\endgroup$
    – cswinson
    Mar 18, 2013 at 20:41

1 Answer 1


I'll go with the definition you mention. Defining $\liminf x_n$ as the smallest limit of all converging subsequences in $[-\infty,+\infty)$ and $\limsup$ as the greatest etc... would make the statement slightly easier to check.

Assume you have proved that $-\sup (-A)=\inf A$ for every subset $A$ of $\mathbb{R}$. Applying this to the set $A=\{x_n\;;\;n>N\}$ yields: $$ -\sup_{n>N}(-x_n)=\inf_{n>N}x_n\qquad \forall N. $$ Taking the limit on both sides gives the formula. Even in the case where these sequences are constant equal to $-\infty$, that is $\liminf x_n=-\infty$ and $\limsup (-x_n)=+\infty$.

Now let us prove the set property. Clearly, $A$ is not bounded below if and only if $-A$ is not bounded above. In this case, we get $-(+\infty)=-\infty$ and the property holds. Now assume we are not in the latter case. Take $a\in A$. We have $$ m\leq a\qquad\iff\qquad -a\leq -m. $$ Therefore, $m$ is a lower bound for $A$ if and only if $-m$ is an upper bound for $-A$. It follows that $-\inf A$ is an upper bound for $-A$, so $\sup(-A)\leq -\inf A$. And likewise, $-\sup(-A)$ is a lower bound for $A$, so $-\sup(-A)\leq \inf A$. This proves the equality $-\sup(-A)=\inf A$.

  • $\begingroup$ Is it safe to take the limit on both sides in the 3rd paragraph? I mean, is there something behind that needs to be proven? $\endgroup$
    – Mangostino
    Feb 13, 2022 at 23:55

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