How to prove the cauchy principle of convergence

The Cauchy principle of convergence states that a sequence converges if and only if it is a Cauchy sequence. To prove this principle, we need to show two things:

1. If a sequence converges, then it is a Cauchy sequence
2. If a sequence is a Cauchy sequence, then it converges

Proof of 1:

Let {an} be a convergent sequence. Then there exists a limit L such that for any ε > 0, there exists an N such that for all n > N, |an - L| < ε.

Now let ε > 0 be arbitrary. We want to show that there exists an N such that for all n, m > N, |an - am| < ε.

Since {an} converges to L, we can choose N1 such that for all n > N1, |an - L| < ε/2.

Similarly, we can choose N2 such that for all m > N2, |am - L| < ε/2.

Let N = max{N1, N2}. Then for all n, m > N,

|an - am| = |an - L + L - am| ≤ |an - L| + |L - am| < ε/2 + ε/2 = ε

Thus, {an} is a Cauchy sequence.

Proof of 2:

Let {an} be a Cauchy sequence. Then for any ε > 0, there exists an N such that for all n, m > N, |an - am| < ε.

Since {an} is Cauchy, it is also bounded. Let M be an upper bound on {an}.

Now consider the set {an : n ≥ N}. This set is bounded above by M and bounded below by {an - ε : n ≥ N}, since for any n ≥ N,

an - ε ≤ an ≤ an + ε

Therefore, by the completeness axiom of the real numbers, this set has a supremum, which we will call L.

We claim that {an} converges to L. To prove this, let ε > 0 be arbitrary. Since {an} is Cauchy, there exists an N such that for all n, m > N, |an - am| < ε/2.

Choose N such that N > N1 and N > N2, where N1 and N2 are the indices corresponding to ε/2 and ε/4, respectively.

Then for all n > N,

|an - L| ≤ |an - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4

Similarly, for all m > N,

|am - L| ≤ |am - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4

Therefore, for all n > N,

|an - L| ≤ |an - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4

and for all m > N,

|am - L| ≤ |am - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4

Thus, for all n > N,

|an - L| < 3ε/4

and for all m > N,

|am - L| < 3ε/4

Therefore, for all n > N,

|an - L| + |L - am| < ε/2 + ε/2 = ε

which shows that {an} converges to L.