Let $(X,d)$ a metric space, $f$ is a function from $X$ to $\mathbb{R}$, and $(x_n)$ is a sequence which converges to $a$. How to proof this inequality? $$\liminf_{x \to a} f(x)\leq \liminf_{n\to\infty} f(x_n)$$
I tried to find it in some books, but i couldn't find any explanation about that inequality and definition of $\liminf_{n\to\infty} f(x_n)$.
Any help will be so appreciated. Thank you.
Suppose the contrary, i.e. $\liminf_{x \to c} f(x)> \liminf_{n\to \infty} f(x_n)$, then, there exists $k\in \mathbb{R}$, such that $\liminf_{x \to c} f(x)> k > \liminf_{n\to \infty} f(x_n)$.
Let $\varepsilon>0$. Since $\liminf_{x \to c}f(x)>k$, it follows that \begin{align*} &\sup_{\varepsilon>0}\{\inf\{f(x):x\in B(c,\varepsilon)\cap X -\{c\}\}\}>k,\\ \Rightarrow & \inf\{f(x):x\in B(c,\varepsilon)\cap X -\{c\}\}>k,\\ \Rightarrow & f(x)>k, \quad \text{for all}\,x\in B(c,\varepsilon). \end{align*} Then, since $(x_n)$ converges to $c$, there exists $n_0\in\mathbb{N}$, such that for all $n\geq n_0$ we have $x_n\in B(c,\varepsilon)$. So, for all $n\geq n_0$, we have $f(x_n)>k$. Therefore, for all $\varepsilon>0$, $\inf \{f(x_n):x_n\in B(c,\varepsilon),n\geq n_0\}\geq k.$ Thus, $$\liminf_{n\to \infty} f(x_n)=\sup\{\inf \{f(x_n):x_n\in B(c,\varepsilon),n\geq n_0\}\}\geq k.$$ It is a contradiction with $k > \liminf_{n\to \infty} f(x_n)$, so we must have $\liminf_{x \to c} f(x)\leq \liminf_{n\to \infty} f(x_n).$