Consequences of convergence

This section gives some consequences of a sequence having a limit. These results are useful sometimes to show that a sequence does not converge.

Proposition 3   Every convergent sequence is bounded.

PROOF:

Suppose that $a_n\to L$ is a convergent sequence. Then there is a N such that if n>N we have |a_n-L|<1 thus there are at most N terms not in the interval [L-1,L+1]. The sequence is then bounded above by the largest element of $\{a_k\vert k\leq N\}\cup \{L-1,L+1\}$ and bounded below by its smallest element.

Because of this proposition we know that if a sequence is not bounded ten it does not converge. We can also show that a sequence does not converge by demonstrating two subsequences which tend to different values.

Definition 3   A sequence b is a subsequence of a if there is a strictly increasing function $f:\ensuremath{\mathbb{N} }\to \ensuremath{\mathbb{N} }$ with b_k=a_f(k).

A subsequence is thus a sequence with some of the terms left out.

Proposition 4   Every subsequence of a convergent sequence converges to the same limit.

PROOF:

Suppose $a_n\to L$ and b is a subsequence of a, say using $f:\ensuremath{\mathbb{N} }\to \ensuremath{\mathbb{N} }$. Given $\epsilon$ there is an N such that if n>N then $\vert a_n - L\vert < \epsilon$. Let M be the smallest natural number such that $f(M)\geq N$ then if m>M we have $\vert b_m-L\vert=\vert a_{f(m)}-L\vert<\epsilon$. Thus $b_k\to L$ as well.

Proposition 5   Every sequence has a monotone subsequence.

PROOF:

If $\lim_{n\to\infty}a_n=\pm \infty$ the proposition is easy: suppose the limit is $\infty$. Pick b₀=a₀. Next let f(n) be the least m with a_m>b_n-1. The resulting subsequence is monotone increasing. A similar argument produces a monotone decreasing subsequence if the limit is $-\infty$. Note that if a sequence a_n is unbounded, then it will have a subsequence which converges to either $\pm\infty$.

Suppose $\{a_n\}$ is bounded. Let $U=\{n\vert\mbox{ for an infinite number of }m>n, a_m\geq a_n\}$ and $L=\{n\vert\mbox{ for an infinite number of }m>n, a_m\leq a_n\}$. Then $U\cup L=\ensuremath{\mathbb{N} }$ since there are an infinite number of natural numbers $m\geq n$ for each n, and thus either $n\in L$ or $n\in U$ or both. Thus at least one of U and L must be infinite.

Suppose U is infinite. Then it is nonempty and we have assumed it is bounded above, so it must have a least upper bound u. We construct a monotone nondecreasing subsequence of a_n converging to u. Let b₀ be the first a_k with $k\in U$. We define f(k+1)=m where m is the smallest element of U larger than f(k) which has $u-a_m <\frac1{k+1}$. This gives us a monotone nondecreasing subsequence of a_n.

A similar argument using the greatest lower bound of L to produce a monotone nonincreasing sequence will work if it is L which is infinite.