2
$\begingroup$

I'm trying to teach myself some basic topology by self-studying from Intro to topology by Mendelson. I'm stuck on one of the exercises and can't figure out how to proceed with the proof.

The question is as below:

Let $a_1,a_2,...$ be a bounded sequence of real numbers. Since each of the sets $A_n =\{ a_n,a_{n+1},...\}$ is bounded, we may set $v_n = g.l.b. A_n$, $u_n = l.u.b. A_n$. Observe that $v_n \leq u_n$; $v_1,v_2,...$ is monotone non-decreasing and bounded above; and $u_1,u_2,...$ is monotone non-increasing and bounded below. Let $V=\lim_{n}\ v_n$ and $U=\lim_{n}\ u_n$.

Prove that there are subsequences of $a_1, a_2,....$ which converge to $U$ and $V$ respectively (thus a bounded sequence of real numbers has a convergent subsequence). Prove that $a_1,a_2,...$ converges if and only if $U=V$.

Any hints/tips would be much appreciated.

Thanks

$\endgroup$
1
$\begingroup$

Note that

$$U = \lim_{n \rightarrow \infty}u_n = \lim_{n \rightarrow \infty}\sup_{k\geq n}a_k,\\V = \lim_{n \rightarrow \infty}v_n = \lim_{n \rightarrow \infty}\inf_{k\geq n}a_k.$$

These are definitions of $\limsup a_n$ and $\liminf a_n.$

Since $u_n$ is non-increasing,

$$\lim_{n \rightarrow \infty}\sup_{k\geq n}a_k= \inf_{n}\sup_{k\geq n}a_k$$

For any $\epsilon >0$ there exists $N \in \mathbf{N}$ such that for $n \geq N$

$$a_n < U + \epsilon.$$

Since $U-\epsilon < \sup_{k\geq m}a_k$, for each $m \geq N$, there exists $k_m \geq m$ such that

$$U - \epsilon < a_{k_m}.$$

Hence, $|a_{k_m} - U| < \epsilon$ when $m \geq N$ and the subsequence $(a_{k_m}$) converges to $U$.

Make a similar argument for $V$.

If $U = V=L$ then there exist $N_U,N_V \in \mathbf{N}$ such that if $n \geq \max(N_U,N_V)$

$$L-\epsilon= V-\epsilon < a_n < U + \epsilon= L + \epsilon,$$

and $a_n$ converges to $L$.

$\endgroup$
  • $\begingroup$ I was trying separately the cases where (i) $u_n \in A_n$ for all $n$ and (ii) where there is some $n$ for which this fails. Your proof combines the two approaches and is much more satisfying. $\endgroup$ – Shai Aug 1 '14 at 22:17
  • $\begingroup$ Thanks. Just have to show convergence implies U equals V which is straightforward . $\endgroup$ – RRL Aug 1 '14 at 23:51

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.