A sequence (a_n) = a_1, a_2, a_3, \dots is an
infinite list of reals. The
idea of a limit
becomes razor-sharp here. We say (a_n) converges
to L, written a_n \to L, when the terms
get — and stay — arbitrarily close to L. Formally:
\forall\, \varepsilon > 0\ \ \exists\, N \in \mathbb{N}\ \ \text{such that}\ \ n \ge N \ \Longrightarrow\ |a_n - L| < \varepsilon.
Read it as a game: you name a tolerance \varepsilon (how close is
"close enough"); I must produce a cutoff N such that
every term from a_N onward sits within
\varepsilon of L. If I can always win,
no matter how tiny your \varepsilon, then
a_n \to L. A sequence is bounded if all its terms
fit in some band: |a_n| \le M for a single
M and all n.
Convergent \Rightarrow bounded
Our first theorem links the two notions: if a sequence converges, it cannot run off to
infinity. The proof is a clean ε–N argument and a model of the genre — a tail controlled
by the limit, a head controlled by finiteness.
Step 1 — pin the tail with one specific \varepsilon.
Convergence holds for every \varepsilon, so we are free to
pick a convenient one. Take \varepsilon = 1. Then there is an
N with
n \ge N \ \Longrightarrow\ |a_n - L| < 1.
Step 2 — bound the tail. For those n \ge N, the
triangle inequality turns closeness-to-L into a bound on
|a_n| itself:
|a_n| = |(a_n - L) + L| \le |a_n - L| + |L| < 1 + |L|.
So every term from a_N onward is bounded by
1 + |L|. The infinite tail is under control.
Step 3 — handle the finite head. Only finitely many terms remain:
a_1, a_2, \dots, a_{N-1}. A finite list of numbers always
has a largest absolute value — call it
K = \max\{\, |a_1|,\ |a_2|,\ \dots,\ |a_{N-1}| \,\}.
Step 4 — combine head and tail. Take the larger of the two bounds:
M = \max\{\, K,\ 1 + |L| \,\}.
Every term is now covered — the head by K, the tail by
1 + |L| — so |a_n| \le M for
all n. The sequence is bounded. \blacksquare
The converse fails: a_n = (-1)^n is bounded
(|a_n| = 1) but oscillates between
-1 and 1 forever, never settling. So
bounded does not force convergence — but it does force the next, deeper, fact.
Bolzano–Weierstrass
A subsequence picks out infinitely many terms in order:
a_{n_1}, a_{n_2}, \dots with
n_1 < n_2 < \cdots. The
Bolzano–Weierstrass theorem says a bounded sequence, even one that never
converges, must contain a convergent subsequence. Here is the bisection sketch.
Step 1 — trap the sequence in an interval. Boundedness puts every term in a
closed interval I_0 = [-M, M].
Step 2 — bisect, keeping the busy half. Cut
I_0 in half. At least one half contains infinitely many terms of
the sequence (two halves cannot both hold only finitely many, or the whole would be
finite). Call that half I_1; its length is
M.
Step 3 — repeat forever. Bisect I_1, again keep a
half with infinitely many terms, getting I_2 of length
M/2; and so on. This yields nested closed intervals
I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots, \qquad \text{length}(I_k) = \frac{2M}{2^k} \to 0.
Step 4 — completeness delivers the point. By the
completeness
of \mathbb{R} (nested closed intervals whose lengths shrink to
0 meet in exactly one point), there is a unique
L \in \bigcap_k I_k.
Step 5 — extract the subsequence. From each
I_k pick a term a_{n_k} with
n_1 < n_2 < \cdots (possible since each
I_k holds infinitely many). Since
a_{n_k} and L both lie in
I_k, whose length \to 0, we get
|a_{n_k} - L| \to 0. So
a_{n_k} \to L: a convergent subsequence.
\blacksquare
-
Convergent \Rightarrow bounded. If
a_n \to L then there is an M with
|a_n| \le M for all n. The converse
is false ((-1)^n is bounded yet diverges).
-
Bolzano–Weierstrass. Every bounded sequence of reals has a
convergent subsequence. Its limit is a limit point of the
sequence, and the theorem rests on the completeness of
\mathbb{R}.