1
$\begingroup$

Given a sequence $\{x_n\}$ of real numbers, I need to show that if $\limsup \{x_k:k\ge n\} = \infty$, then sequence $x_n$ is unbounded above and that $\infty$ is the partial limit (subsequence limit) of $x_n$.

I can intuitively see why the result hold but I'm having a hard time showing it rigorously. So far my proof looks as follows;

Define $a_n=\sup\{x_k:k\geq n\}$, then $a_n$ is a decreasing sequence and since $\{a_n\}\to\infty$ , $\{a_n\}$ must be equal to $\infty$ for all $n$.

I'm stuck here, I don't know how can I make the argument about $\{x_n\}$ being unbounded from here. Any ideas?

Also how can I show that $\infty $ is the subsequence limit.

$\endgroup$
1
  • $\begingroup$ Have you tried proving this by contradiction? $\endgroup$
    – Yunus Syed
    Mar 5 '17 at 23:57
2
$\begingroup$

The definition of bounded above is having an upper bound that is a real number. A nonempty set $A$ that is bounded above has a least upper bound $\sup(A)$, which is a real number. If $A$ is not bounded above we say $\sup(A)=\infty$.

You observed that the hypothesis implies that $\sup\{x_k:k\geq n\}=\infty$ for all $n$. By definition, $\sup\{x_k:k\geq n\}=\infty$ means $\{x_k:k\geq n\}$ is not bounded above. Because this is a set of terms from the sequence, this implies the sequence is not bounded above (by definition).

To construct a subsequence diverging to $\infty$, you can use the fact that each $\{x_k:k\geq n\}$ is unbounded above as follows:

Because $\{x_k:k\geq 1\}$ is ubounded above, take $k_1\geq 1$ such that $x_{k_1}>1$. Because $\{x_k:k\geq k_1+1\}$ is unbounded above, take $k_2>k_1$ such that $x_{k_2}>2$. In general, given $k_1<k_2<\cdots<k_{n-1}$, because $\{x_k:k\geq k_{n-1}+1\}$ is unbounded above, there is a $k_n>k_{n-1}$ such that $x_{k_{n}}>n$. This yields a subsequence $(x_{k_n})$ such that $x_{k_n}>n$ for all $n$, from which it follows that $\lim\limits_{n\to\infty}x_{k_n}=\infty$.

$\endgroup$

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.