Prove that there exists a sequence $\{x_n\}_{n \in \mathbb{N}} \subseteq A$ such that $\lim_{n \to \infty} x_n = I$

Question: Let $A \subseteq \mathbb{R}$ where $A$ is non-empty and bounded. Assume $I = \inf(A)$. Prove that there exists a sequence $\{x_n\}_{n \in \mathbb{N}} \subseteq A$ such that $\lim_{n \to \infty} x_n = I$.

I thought about (1) using the definitions: $A$ is bounded $\iff \exists h\in \mathbb{R} : |a|\leq h \; \forall a \in A$ and $I = \inf(A) \iff \forall a \in A, \; I \leq x \text{ and } \forall \varepsilon>0, \; \exists a \in A: x<I+\varepsilon$;

and (2) to define a constant sequence that is constant and only has $I$ in it (if $I \in A$). But idk what to do when $I \notin A$.

and kinda got nowhere in both cases. Can someone give me a hint?

Solution 1:

Hint: If $I = \inf A$ then in particular as you say for all $\varepsilon > 0$ there exists $a_\varepsilon \in A$ such that $a_\varepsilon < I + \varepsilon$. (The only thing boundedness gets us is finiteness of $I$.) Now set $x_n := a_{\frac{1}{n}}$. (Clearly the sequence $(x_n)$ is contained in $A$.) Can you show that it converges (essentially by definition) to $I$?

Spoiler: Indeed $x_n \to I$, since by construction $\lvert x_n - I \rvert = x_n - I < \frac{1}{n}$ for all $n$.