If $l$=sup($x_n$), what is sup($kx_n$) where k$\in$$\mathbb{R}^{+}$? Prove your conjecture.
I have that sup($kx_n$)=$kl$. I can prove that it is an upper bound of $kx_n$, but I'm having trouble finishing the proof and showing that it is less than or equal to any other upper bound of $kx_n$.
On
To show that $kl$ is the least upper bound, suppose, for a contradiction, that you can find a smaller lower bound. That is, suppose there exists $b \in \mathbb{R}$ with $b < kl$ and $kx_n \le b$ for all $n\in\mathbb{N}$. So for all $n\in\mathbb{N}$,
$$ kx_n \le b < kl. $$
But then $$ x_n \le \frac{b}{k} < l $$ for all $n\in\mathbb{N}$.
Do you see the contradiction?
Hint: Given $\varepsilon > 0$ and $x_n$ with $x_n>l - \varepsilon/k$, $k x_n > kl - k \varepsilon/k = kl - \varepsilon$. Why does this $x_n$ exist, and why is it enough to get your result?