Subsequence

Suppose $u:\mathbb{Z}^+\rightarrow \mathbb{R}$ be a sequence and $v:\mathbb{Z}^+\rightarrow\mathbb{Z}^+$ be an increasing sequence. Then $u\circ v: \mathbb{Z}^+\rightarrow \mathbb{R}$ is a subsequence of $u$.

Monotonic subsubsequence

Every sequence has a monotonic subsequence.

Proof

• Let $n\in\mathbb{Z}^+$ be called “good” iff $\forall m>n,\,u_n > u_m$.
• Suppose $u_n$ has infinitely many “good” points. That implies $u_n$ has a decreasing subsequence.
• Suppose $u_n$ has finitely many “good” points. Let $N$ is the maximum of those. $\forall n_1 > N,\,n_1\;\text{is not "good"}$ That implies $u_n$ has a increasing subsequence.

Bolzano-Weierstrass

Every bounded sequence on $\mathbb{R}$ has a converging subsequence. $$

Proof

From the above theorem, there is a monotonic subsequence $u_{n_k}$ which is also bounded. Bounded monotone sequences converge. $$

Note

For a set $A$, all $3$ statements are equivalent:

• $A$ has the completeness property
• $A$ is complete
• Bolzano-Weierstrass theorem on $A$

Convergence

Suppose $u_{n_k}$ is a subsequence of $u_n$.

Sequence converging

$$\lim_{n \to \infty} u_n = L \implies \lim_{n_k \to \infty} u_{n_k} = L$$

Sequence diverging to infinity

$$\lim_{n \to \infty} u_k = \infty \implies \lim_{n_k \to \infty} u_{n_k} = \infty$$

Proof Hint

Above 2 conclusions can be derived using $n_k \ge k$. $$

Converging subsequence

If $u_n$ is Cauchy and $u_{n_k}$ is a subsequence converging to $L$, then $u_n$ converges to $L$.