0

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 $kx_n$.

Lauralolo
  • 91

2 Answers2

1

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?

Ian
  • 101,645
  • Would the downvoter care to comment? – Ian Mar 30 '15 at 00:21
  • I'm confused as to how the $\epsilon$ helps in proving that another upper bound, say $p$, of $kx_n$ will be greater or equal to $kl$ – Lauralolo Mar 30 '15 at 00:26
  • @Lauralolo You need to prove two things: that $kl$ is an upper bound, and that it is the least one. You said you have already proven it is an upper bound. My hint will help you show it is the least one. – Ian Mar 30 '15 at 00:32
  • @Lauralolo What I am attempting to show here is that there are numbers in $k x_n$ which are arbitrarily close to $kl$. From this it follows that no number can be a smaller upper bound. – Ian Mar 30 '15 at 00:41
0

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?

sardoj
  • 471