A sequence is bounded if and only if there is a $C > 0$ such that $|x_n| \leq C$ for all $n$ Suppose that $x_n$  is a sequence of real numbers. 
Prove that $\{x_n\}$ is bounded if and only if there is a $C > 0$ such that $|x_n| \leq C$ for all $n \in \mathbb{N}$.
Attempt: (proof →) Suppose $\{x_n\}$ is a sequence of real numbers. And suppose $\{x_n\}$ is bounded. Then by definitions a sequence is bounded iff it is bounded both above and below.
Then, there exixts $C$ in $\mathbb{R}$ such that $x_n < C$ for all $n$ in $\mathbb{N}$. Likewise, $x_n > -C$, for all $n$ in $\mathbb N$. Thus, $|x_n| \leq C$, for all $n$ in $\mathbb{N}$.
Can anyone please give feedback? I don't know if this makes sense. Thank you.
 A: You seem to be using the following definitions. 
A sequence of real numbers $\{x_n\}$ is bounded below if there is a real number $A$ such that $A < x_n$ for all $n$, and is bounded above if there is a real number $B$ such that $x_n < B$ for all $n$; we often call $A$ and $B$ lower and upper bounds respecitively. If a sequence is bounded above and below, it is said to be bounded.
Turning to your proof, you said that as $\{x_n\}$ is bounded above, there is $C$ such that $x_n < C$ for all $n$. This is true, however, you claim that as $\{x_n\}$ is bounded below, $-C < x_n$ for all $n$; this is false. As $x_n$ is bounded below, there is some real number $A$ such that $A < x_n$, but $A$ may or may not be $-C$. For example, if $x_n = -\frac{2}{n}$, then $x_n < 1$ for every $n$, but $-1 < x_n$ does not hold for every $n$.
To prove the claim, you have to work with a lower bound $A$ and an upper bound $B$, which exist because $\{x_n\}$ is bounded, then decide how to choose $C$ appropriately. One thing to be weary of is that $A$ is not necessarily negative and $B$ is not necessarily positive.
The reverse implication should be simpler. Using the condition $|x_n| \leq C$, you should be able to find a lower bound and an upper bound.
