# Set $S={x^{3}+2x<4}$ bounded from above or below?

It's obvious that it won't be bound below and the LUB will be ~$1.18$, but how do you prove that it is indeed bounded from above?

If you assume that it is not bounded from above, then, by definition this means that $\forall n\in\mathbb N$ there is $x_n\in S:x_n>n$. Then you have a sequence $\left\{x_n\right\}\subset S$ such that $x_n\to\infty$. But for this sequence $x_n^3+2x_n\to\infty$ which is a contradiction with that $x_n^3+2x_n<4$. Therefore $S$ is bounded !

• I don't see what the assumption has to do with anything. Commented Sep 17, 2015 at 18:06
• @pjs36 If it is not bounded from above , by definition this means that $\forall n\in\mathbb N$ there is $x_n\in S:x_n>n$. Then you have a sequence $\left\{x_n\right\}\subset S$ such that $x_n\to\infty$. But for this sequence $x_n^3+2x_n\to\infty$ which is a contradiction with that $x_n^3+2x_n<4$ Commented Sep 17, 2015 at 18:09
• I'm merely pointing out that it is superfluous to assume such a thing. Your contradiction comes from your claim that "$x_n^3 + 2x_n < 4$," which, if justified, would be enough for a direct proof. Commented Sep 17, 2015 at 18:19
• @pjs36 I can not understand what you have in mind. The set $S$ consists of such real numbers $x$ that $x^3+2x<4$. What is your concern here ? Commented Sep 17, 2015 at 18:23
• Yes, I was clearly confused there, you're exactly right. My apologies! Commented Sep 17, 2015 at 18:35

You could look at the function $f:\mathbb{R}\to \mathbb{R}, \ x\mapsto x^3+2x$, the derivative is $3x^2+2$ which is positive everywhere, hence your function is strictly monotone and $f(2)=12>4$ which means 2 is an upper bound for your set $S$.

If $x \ge 2$, then $x^3 + 2x \ge 2^3+2 \cdot 2 =12$.

Thus, if $x^3 + 2x < 4 < 12$, then $x < 2$. So, $2$ is an upper bound for $S$.

If $x<0$, then $x^3+2x < 0 < 4$. So $S$ contains all negative numbers and is not bounded below.

• Is it possible to say it isn't bounded below? Commented Sep 17, 2015 at 18:05
• Just so I don't have to open another question that is very related to this one. If $A=r^{2}<9$, $inf(A)=-3$ $sup(A)=3$, would something like this need to be proved? Commented Sep 17, 2015 at 18:16
• @Achro, yes. Try it. If you get stuck, ask a separate question.
– lhf
Commented Sep 17, 2015 at 18:17