Let $A = \{x\in \mathbb{R}: a_n < x \text{ for infinitely many n}\}.$ Prove that $\inf A = \liminf _{n\to \infty} a_n$. Let $(a_n)^{\infty} _{n=1}$ be a bounded sequence of real numbers. Let $$A = \{x\in \mathbb{R}: a_n < x \text{ for infinitely many n}\}.$$ Prove that $\inf A = \liminf _{n\to \infty} a_n$.
I think I understand intuitively. If they are not equal, then $A$ would not contain infinitely many terms of $a_n$, right? But I'm kind of clueless on how to prove this rigorously.
 A: Let $\alpha = \inf A$ and $\beta = \liminf a_n$. You want to show that $\alpha \leq \beta$ and $\beta\leq \alpha$.
$\beta\leq \alpha$ : For each $n$, write
$$
u_n = \inf\{a_n, a_{n+1}, a_{n+2}, \ldots\}
$$
For each $x\in A$, $a_n < x$ for infinitely many $n$. Hence, there is some $k > n$ such that $a_k < x$. Thus $u_n < x$. Thus,
$$
\beta = \sup u_n \leq x
$$
This is true for all $x \in A$, and hence $\beta \leq \inf A = \alpha$.
$\alpha \leq \beta$ : For any $\epsilon >0$, $\beta + \epsilon > u_n$ for all $n$. Thus, for $n\in \mathbb{N}$ fixed, $\beta +\epsilon$ is not a lower bound for $\{a_n, a_{n+1}, a_{n+2}, \ldots\}$. Hence, there is an $a_{n + k_n} < \beta+\epsilon$. This is true for all $n\in \mathbb{N}$, and hence there are infinitely many $a_n$ such that $\beta + \epsilon > a_n$.
Thus $\beta + \epsilon \in A$; and so $\beta + \epsilon \geq \alpha$. This is true for all $\epsilon > 0$, and hence, $\beta \geq \alpha$.
A: Let  be $a = \inf A$ and $b = \liminf_{n\to \infty} a_n$.
If $a< b$, then there is $\varepsilon > 0$, such that $a<b-\varepsilon $. It follows from the definiton of $b$ that there are only finitely many terms in interval $I=(a-\frac{\varepsilon}{2},a+\frac{\varepsilon}{2})$, since $b\notin I$. Therefore $I\cap A = \emptyset$, so $\inf A\notin I$ (Why?) and we come to contradiction.
If $a>b$, then there is $\varepsilon>0$ such that $a>b+\varepsilon$. There are infinitely many terms $a_n$ in the interval  $J = (b-\frac{\varepsilon}{2},b+\frac{\varepsilon}{2})$, therefore $a' = b+\frac{\varepsilon}{2}\in A$. It follows that $a'<a$ and again, we have contradiction, so $a=b$.
