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.



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 $L$.

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  • $\begingroup$ 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

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