Relationship between Limit Inferior and Lower Limit

Theorem
Let $\left({S, \tau}\right)$ be a topological space.

Let $f: S \to \R \cup \left\{{-\infty, \infty}\right\}$ be an extended real-valued function.

Let $\langle s_n \rangle_{n \in \N}$ be a convergent sequence in $S$ such that $s_n \to \bar s$.

Then the lower limit of $f$ at $\bar s$ is bounded from above by the limit inferior of $\langle f(s_n) \rangle$:


 * $\displaystyle \liminf_{s \mathop \to \bar s} f \left({s}\right) \leq \liminf_{n \mathop \to \infty} f \left({s_n}\right)$

Proof
By definition of the lower limit, there exists a sequence of open neighborhoods $\langle V_k \rangle_{k \in \N} \in \mho \left({ \bar s}\right)$ such that


 * $\displaystyle  \lim_{k \mathop \to \infty} \left\{ \inf_{s \mathop \in V_k} f \left({s}\right) \right\} = \liminf_{s \mathop \to \bar s} f \left({s}\right)$

This implies that $\forall \varepsilon > 0$ $\exists k_\varepsilon \in \N$ such that


 * $\displaystyle  \inf_{s \mathop \in V_{k_\varepsilon}} f \left({s}\right) \geq \liminf_{s \mathop \to \bar s} f \left({s}\right) - \varepsilon$

By our hypothesis $s_n \to \bar s$ and because $V_k \in \mho \left({ \bar s}\right)$ there exists $N \left({k_\varepsilon}\right) \in \N$ such that


 * $\displaystyle \forall n \geq N \left({k_\varepsilon}\right) : s_n \in V_{k_\varepsilon}$

Consequently


 * $\displaystyle  \inf_{s \mathop \in V_{k_\varepsilon}} f \left({s}\right) \leq \inf_{n \geq N \left({k_\varepsilon}\right)} f \left({s_n}\right) \leq \sup_{N \in \N} \left\{ \inf_{n \geq N} f \left({s_n}\right) \right\} = \liminf_{n \mathop \to \infty} f \left({s_n}\right)$

Combining these estimates, it follows that for all $\varepsilon > 0$


 * $\displaystyle \liminf_{s \mathop \to \bar s} f \left({s}\right) \leq \liminf_{n \mathop \to \infty} f \left({s_n}\right) + \varepsilon$

Hence the result.