If $x_n \ge y_n$ and $y_n$ is not bounded above then $x_n$ is not bounded above I'm not quite sure how to prove this property of limits with sequences, can someone help?

Show that if $x_n ≥ y_n$ and $y_n$ is not bounded above then $x_n$ is not bounded above.
I'm supposing that $(x_n), (y_n)$ are sequences in $\mathbf{R}$. Suppose that for all $n\in \mathbf{N}$, $x_n\ge y_n$, and $y_n$ is not bounded above. That is, there exists no $M\in \mathbf{R}$ such that $M\ge y_n$ for all $n\in \mathbf{N}$. If there existed an $N\in \mathbf{R}$ with the property that $N\ge x_n$ (an upper bound) for all $n\in \mathbf{N}$, then this would imply that $N\ge y_n$ for all $n\in \mathbf{N}$ - contrary to assumption.
You can say that a sequence is not bounded from above, if $\exists n,\not\exists M \in \mathbb{R} : y \le M_n$, just the negation of the definition of upper bound, but you can also say that "no matter what number you give me, there's always (at some place; with some value of $n$, an element of the sequence $y_n$ which is larger than the number you give me - so the sequence it's not bounded from above, because you'd be going over all the numbers (in $\mathbb{R}$):

$y_n$ is not bounded from above $\equiv \forall n,\exists M \in \mathbb{R} 
: y_n &gt;M$.

$ \Rightarrow \forall n,\exists M\in\mathbb{R}: x_n \ge y_n &gt; M \Rightarrow x_n &gt; M \Rightarrow x_n$ is not bounded from above.

$\square$