Show that the lim inf $\{a_n\}$ is the smallest cluster point Show that the lim inf $\{a_n\}$ is the smallest cluster point. 
I need to show 
1) The lim inf $\{a_n\}$ is a cluster point exist
2) Show that the lim inf $\{a_n\}$ is the smallest cluster point 
proof: (1)
For all $n$ we can find some $\{a_l\}$ where $l_n \geq n$ such that inf $\{a_n, a_{n+1}...\}$ $<$ $a_{l_{n}} + \frac {1}{n}$. the $a_{l_{n}}$ = lim inf$\{a_n\}$. 
i do not know if that is correct, and also i have no idea of how to show the second part 

Let's consider the sequence:
$$b_{1} = \inf \left \{ a_{n} \right \}_{n=1}^{\infty }$$
$$b_{2} = \inf \left \{ a_{n} \right \}_{n=2}^{\infty }$$
$$...$$
$$b_{k} = \inf \left \{ a_{n} \right \}_{n=k}^{\infty }$$
So
$$\lim \inf \left \{ a_{n} \right \} = \lim_{k \to \infty } \left ( \inf \left \{ a_{n} \right \}_{n=k}^{\infty } \right ) = \lim_{k \to \infty } b_{k}=b$$
And $$b_{1} \leqslant  b_{2} \leqslant ... \leqslant b_{k} \leqslant ...$$
From the definition of $\inf$, $\forall \varepsilon >0$:
$$\exists a_{n_{1}}\in \left \{ a_{n} \right \}_{n=1}^{\infty }: b_{1} \leq a_{n_{1}} < b_{1} + \varepsilon $$
$$\exists a_{n_{2}}\in \left \{ a_{n} \right \}_{n=2}^{\infty }: b_{2} \leq a_{n_{2}} < b_{2} + \varepsilon $$
$$...$$
$$\exists a_{n_{k}}\in \left \{ a_{n} \right \}_{n=k}^{\infty }: b_{k} \leq a_{n_{k}} < b_{k} + \varepsilon $$
And $$\left | a_{n_{k}} - b \right | = \left | (a_{n_{k}} - b_{k}) + (b_{k} - b) \right |\leq \left | a_{n_{k}} - b_{k} \right | + \left |b_{k} - b \right | \leq 2\cdot \varepsilon$$
So, $b$ has infinity of $\left \{ a_{n} \right \}_{n=1}^{\infty }$ in any of its vicinity. Thus, it's a cluster point. I excluded the $b=\infty $ case, but similar technique can be applied.
Re part 2, try proving by contradiction. Let's assume $a$ is another cluster point such that $$a < b$$
Because 
$$\lim_{k \to \infty } \left ( \inf \left \{ a_{n} \right \}_{n=k}^{\infty } \right ) =b$$ 
this means that $\forall \varepsilon >0,\exists n_{0}\in \mathbb{N}:\forall n> n_{0}$ $$b-\varepsilon < a_{n}$$ and we can choose $\varepsilon$ such that $$a < a + \varepsilon < b - \varepsilon < b$$
This means that $\left ( a-\varepsilon , a+\varepsilon  \right )$ will contain only a finite subsequence of $\left \{ a_{n} \right \}_{n=1}^{\infty }$, i.e. none, some or all of $\left \{ a_{n} \right \}_{n=1}^{n_{0} }$. So $a$ is not a cluster point.
