Proving that sequence is bounded Given $|x_n| \leq y_n$ for all $n \in N$, and $(y_n)$ is a convergent sequence, prove that $(x_n)$ is bounded as well. 

My attempt:
Let, $\lim(y_n) = a,$ then since $|x_n| \leq y_n$ for all $n,$ we have $|x_n| \leq a.$ Hence, ($|x_n|$) is bounded by $a.$

How do I now go about proving $(x_n)$ is bounded as well?
The statement

" $\lim(y_n) = a$, then since $∣x_n∣ ≤ y_n$ for all $n$, we have $∣x_n∣ ≤ a$ "

is not necessarily true.

You may alter it to 

"since  $\lim(y_ n) = a$, we have $ y_n \le M$ for some $M&gt;0$ and since $∣x_n∣ ≤ y_n$ for all $n$, we have $∣x_n∣\le M "$ 

We know that $- ∣x_n∣\le  x_n \le ∣x_n∣ $, so $x_n$ is bunded.
Try thinking about how $|x_n|\leq y_n$ is the same as $|x_n-0|\leq y_n$.
As $y_n$ is bounded we know there exist some upper bound $U$, we also know that $y_n \ge0$ as $y_n\ge|x_n|$ so $|y_n|=y_n&lt;U$
And $|x_n|\le y_n &lt;U$
At very least $x_n$ is bounded by $+U,-U$