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.

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

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 $\lim\limits_{n\to\infty}x_{k_n}=\infty$.


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