Set of subsequent limits 
If  a sequence is bounded then set of all subsequential limit of the sequence is bounded. 

Is the converse true?
 A: Yes, it's true.
If the sequence itself is not bounded, then the sequence must have a subsequence that is unbounded which means that the set of subsequential limits of the sequence must contain $+\infty$ or $-\infty$ and that's contradictory.
To show the existence of such a subsequence, go forward like this:
Suppose that $a_n$ is unbounded, therefore for any $M>0$ we have $|a_n| > M$ for some $n \in \mathbb{N}$.
Set $M=1,2,\cdots$ then you find a subsequence $\{a_{n_k}\}$ such that $|a_{n_k}|>n$ and this subsequence is unbounded.
A: It depends on whether you consider $\infty$ and $-\infty$ to be valid "subsequential limits" (which are more commonly called "accumulation points" of a sequence).  I.e., it depends on whether you work in the topological space $\mathbb{R}$ or the topological space $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, \infty\}$.


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*If $\infty$ and $-\infty$ are considered valid accumulation points, then a sequence is bounded if and only if it contains the accumulation point $\infty$ or the accumulation point $-\infty$ (or both).

*If $\infty$ and $-\infty$ are not considered valid accumulation points, then consider the sequence $1, 2, 3, 4, \ldots$.  This sequence has no accumulation points, so the set of accumulation points is bounded, vacuously.  But this sequence is not bounded.
