Let A be a bounded subset of R. Prove that there is an infinite sequence (Xn) such that xn ∈ A for every n, and lim n→∞ xn = inf(A).
My attempt: Let inf(A)= a. Then since a is a greatest lower bound for A, there exists an xn ∈ A such that: a < xn < inf + 1/2. The same is true for inf + 1/3, inf+ 1/4,..., inf + 1/n. This selection of xn's based on 1/n gives a sequence contained in A.
I'm not sure if I my construction of the sequence needs more detail. From here, I need to show this sequnce converges to a using the epsilon definition of a limit, but I'm struggling with manipulating this 'implicitly' described sequence. Any help is appreciated!
Suppose on the contrary that $\exists n$ such that there $[a, a+\frac1n) \cap A =\emptyset$, then $a+\frac1n$ is a lower bound of $A$ which violates $a$ being the greatest lower bound.
Hence we can always choose $x_n \in [a,a+\frac1n]$
Hence by construction, we have $|x_n - a| \le \frac1n$.
Hence $$\lim_{n \to \infty} |x_n-a| \le \lim_{n\to \infty} \frac1n = 0$$
Remark: If you want to use an explicit $\epsilon$ definition of a limit , just choose $N > \frac1{\epsilon}$.