Let $\{x_n\}_{n=0}^\infty$ be a sequence of real numbers. Let  $\{x_n\}_{n=0}^\infty$ be a sequence of real numbers.
 Let $\alpha = \liminf_{n\to\infty} x_n$.
Which of the following statements are true?
a. For every $\epsilon \gt 0$, there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \leq \alpha + \epsilon$
for all $k\in N$.
b. For every  $\epsilon \gt 0$, there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \leq \alpha - \epsilon$
for all $k\in N$.
c. There exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \rightarrow \alpha$  as $k\rightarrow \infty$.
If the sequence converges then
$$\liminf_{n\to\infty} x_n=\limsup_{n\to\infty} x_n$$
then $x_n \lt \alpha + \epsilon$ (only a few will be greater than $b+\epsilon$) then subsequence will also follow and a,c will be right...What if sequence diverges?
In exam I took particular examples like $-\frac {1}{n}$ and $-\frac{1}{2n}$ and such...but don't know proper explanation...
 A: Fact: the set $S$ of subsequential limits of a sequence $(x_n)_n$ is a closed subset of the extended reals $[-\infty, +\infty]$. Its maximum is $\limsup_n x_n$ and its minimum is $\liminf_n x_n$.
This immediately implies that c) holds. 
a) Also holds, for it it failed, there would exists a $\varepsilon_0 > 0$ such that all convergent subsequences have limits $> \alpha + \varepsilon_0$ ,this implies that $S \subseteq (\alpha+\epsilon_0 ,\infty]$ so that $\alpha$ cannot be its minimum, contradiction.
A: (a) is true.  Let's see why using the definition of lim inf.
Let $\epsilon > 0$.  $\liminf \limits_{n \to \infty} x_{n} = a$ means $\lim \limits_{n \to \infty} g_{n} = a$, where $g_{n} := \inf \limits_{k \geq n} \{x_{k}\}$, right?  So, for some point $N$ in the sequence, we have all later points (i.e., for all $n \geq N$) satisfying $g_{n} \in (a - \epsilon, a + \epsilon)$.  That means $\inf \limits_{k \geq n}\{x_{k}\} \in (a - \epsilon, a + \epsilon)$ for all $n \geq N$.  Ok, so let's look at $N$ first.  $g_{N} \in (a - \epsilon, a + \epsilon)$ implies $\inf \limits_{k \geq N} \{x_{k}\} \in (a - \epsilon, a + \epsilon)$, and so by properties of infimum, we can find some point in the sequence $x_{n_{1}}$, with $n_{1} \geq N$.  Now, since $n_{1} \geq N$, that means $g_{n_{1}} \in (a - \epsilon, a + \epsilon)$.  By the same argument as above, we can find some $n_{2} > n_{1}$ so that $x_{n_{2}} \in (a - \epsilon, a + \epsilon)$.  Then we can find $n_{3} > n_{2}$, and so on.  In particular, we have a subsequence $x_{n_{k}}$ satisfying $x_{n_{k}} < a + \epsilon$, which is what we wanted to find, so (a) is true.

(b) is false, as Daniel Schepler noted in the comments.  
To show this using the definitions, note that $a - \epsilon < a$, right?  Now, if $\liminf \limits_{n \to \infty} x_{n} = a$, that means $\lim \limits_{n \to \infty} g_{n} = a$, where $g_{n} := \inf \limits_{k \geq n} \{x_{k}\}$, right?  So, after some $N$, all of the $g_{n}$ are within $\frac{\epsilon}{2}$-distance of $a$ by definition of limit.  In particular, for some $N$, $a - \frac{\epsilon}{2} < g_{N} = \inf \limits_{k \geq N} \{x_{k}\}$, implying that $a - \frac{\epsilon}{2} \leq x_{k}$ for all $k \geq N$.  This shows that you can't find a subsequence less than $a - \epsilon$.
