Prove the infimum of monotone decreasing set equals its limit

Question

Suppose that the sequence (sn) is monotone decreasing, in other words s1 ≥ s2 ≥ .... In addition suppose (sn) converges to s ∈ R. With these assumptions, prove that the set E = {s1, s2, ...} has an infimum and that inf E = s.

Answer 1

Hint: Proceed as follows.

Certainly, $s$ is a lower bound. Assume, however, that $s$ is not the greatest lower bound. Instead, perhaps there exists a larger one, say $s+ \epsilon$ for some $\epsilon > 0$.

Can you arrive at a contradiction? (Apply the definition of convergence).

Answer 2

Since $E$ is bounded below by $s$, $Inf(E) = a$ exists. Let $\epsilon > 0$ be given, there exists $N$ with $s_N < a + \epsilon$, and since $(s_n)$ is a decreasing sequence, $s_N \geq s_n$ for all $n \geq N$, so: $s_n \le s_N < a + \epsilon$, but it is already true that: $s_n \ge a > a - \epsilon$ for all $n$. Thus: for all $n \geq N$, $a - \epsilon < s_n < a + \epsilon$ or $|s_n - a| < \epsilon$, showing that $\displaystyle \lim_{n \to \infty} s_n = a$. so $a = s$