# 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

Let $\mathcal N_{\bar s}$ denote the neighborhood filter of $\bar s$.

By definition of the lower limit, there exists a sequence of open neighborhoods $\langle V_k \rangle_{k \in \N} \in \mathcal N_{\bar s}$ 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 $V_{k_\varepsilon} \in N_{\bar s}$ 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:

 $\displaystyle \inf_{s \mathop \in V_{k_\varepsilon} } f \left({s}\right)$ $\le$ $\displaystyle \inf_{n \geq N \left({k_\varepsilon}\right)} f \left({s_n}\right)$ because $\left\{ s_n ~:~ n \geq N \left({k_\varepsilon}\right) \right\} \subseteq V_{k_\varepsilon}$ $\displaystyle$ $\le$ $\displaystyle \sup_{N \in \N} \left\{ \inf_{n \geq N} f \left({s_n}\right) \right\}$ Definition of supremum. $\displaystyle$ $=$ $\displaystyle \liminf_{n \mathop \to \infty} f \left({s_n}\right)$ Definition of limit inferior

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.

$\blacksquare$