To check
convergence
directly you must already know the limit L — you compare each term
against it. But often the limit is the very thing you are trying to construct. The
Cauchy criterion sidesteps this: it asks only that the terms
bunch together among themselves, with no reference to any limit. A sequence
(a_n) is Cauchy when
\forall\, \varepsilon > 0\ \ \exists\, N \in \mathbb{N}\ \ \text{such that}\ \ m, n \ge N \ \Longrightarrow\ |a_m - a_n| < \varepsilon.
Past the cutoff N, any two terms — not just term-versus-limit
— differ by less than \varepsilon. The whole tail collapses into a
window of width \varepsilon. The headline result of this page is
that in \mathbb{R} this internal bunching is
exactly equivalent to convergence.
Convergent \Rightarrow Cauchy
One direction is easy and holds everywhere. If the terms approach a single limit, they must
approach each other. The engine is the triangle inequality, run with a
\tfrac\varepsilon2 + \tfrac\varepsilon2 split.
Step 1 — convergence, but aimed at \tfrac\varepsilon2.
Suppose a_n \to L. Convergence holds for every tolerance, so apply
it to the half-tolerance \tfrac\varepsilon2: there is an
N with
n \ge N \ \Longrightarrow\ |a_n - L| < \tfrac{\varepsilon}{2}.
Step 2 — compare two tail terms via the limit. Take any
m, n \ge N. Insert and remove L — a zero
— then apply the triangle inequality:
|a_m - a_n| = |(a_m - L) + (L - a_n)| \le |a_m - L| + |a_n - L|.
Step 3 — add the two halves. Both m and
n exceed N, so each term on the right is
below \tfrac\varepsilon2:
|a_m - a_n| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon.
That is the Cauchy condition. So every convergent sequence is Cauchy — and notice we never
used any special feature of \mathbb{R}; this half is true in
\mathbb{Q} too. \blacksquare
Cauchy \Rightarrow convergent (in \mathbb{R})
The reverse is the deep direction, and it is where completeness enters. It
is false in \mathbb{Q} (see the vignette) and true in
\mathbb{R}. The argument runs in three moves.
Step 1 — a Cauchy sequence is bounded. Apply the Cauchy condition with
\varepsilon = 1: past some N, all terms
sit within 1 of a_N, hence inside the
bounded window (a_N - 1,\ a_N + 1); the finitely many earlier terms
are bounded too. So the whole sequence is bounded — the same head/tail argument as for
convergent sequences.
Step 2 — extract a convergent subsequence. Being bounded, by
Bolzano–Weierstrass
the sequence has a subsequence a_{n_k} \to L for some
L \in \mathbb{R}. This is precisely the step completeness powers —
L exists because \mathbb{R} has no gaps.
Step 3 — promote the subsequence to the whole sequence. A Cauchy sequence
with a convergent subsequence converges (to the same limit). Given
\varepsilon > 0, choose N so that
m, n \ge N \Rightarrow |a_m - a_n| < \tfrac\varepsilon2, and pick
a subsequence index n_k \ge N with
|a_{n_k} - L| < \tfrac\varepsilon2. Then for all
n \ge N,
|a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon.
So a_n \to L. The bunching of the tail, anchored to the
subsequence's limit, drags the entire sequence to L.
\blacksquare
-
Always: every convergent sequence is
Cauchy (triangle inequality with
\tfrac\varepsilon2 + \tfrac\varepsilon2). This holds in any
ordered field.
-
In \mathbb{R}: every Cauchy
sequence converges — so in \mathbb{R},
Cauchy \Leftrightarrow convergent. This direction
needs completeness (via Bolzano–Weierstrass) and is exactly what fails in
\mathbb{Q}.
-
A metric space in which every Cauchy sequence converges is called
complete; the theorem says \mathbb{R} is
complete, an equivalent restatement of the least-upper-bound axiom.
Watch the tail bunch within ε
Below is a Cauchy sequence a_n = 2 + \tfrac{\sin n}{n} (it
converges to 2). Drag \varepsilon to set
a window (L - \varepsilon,\ L + \varepsilon); the dashed cutoff
N marks where the tail first enters — and never leaves — that
window. Past N every pair of terms differs by less than
2\varepsilon: the Cauchy condition, on screen. Shrink
\varepsilon and N slides right, but a
cutoff always exists.
The "Cauchy \Rightarrow convergent" direction fails over
\mathbb{Q}, and that failure is the gap we met in
the real number system.
Define rationals by the decimal expansion of \sqrt{2}:
a_1 = 1.4,\quad a_2 = 1.41,\quad a_3 = 1.414,\quad a_4 = 1.4142,\ \dots
Every a_n is rational (a terminating decimal). The sequence is
Cauchy: |a_m - a_n| \le 10^{-\min(m,n)+1} \to 0, so the terms
bunch arbitrarily tightly. Yet its only possible limit is
\sqrt{2}, which is
irrational.
So within \mathbb{Q} the sequence is Cauchy but has
no limit — it converges toward a hole.
This is the cleanest definition of completeness: a space is complete exactly when
Cauchy = convergent. \mathbb{R} is, by design, the
completion of \mathbb{Q} — you can literally
build \mathbb{R} as equivalence classes of Cauchy
sequences of rationals, the construction of
Cauchy and
Dedekind's contemporaries.