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?
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3 Answers
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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 !
answered Sep 17, 2015 at 17:58
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$\begingroup$ I don't see what the assumption has to do with anything. $\endgroup$– pjs36Commented Sep 17, 2015 at 18:06
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1$\begingroup$ @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 $x_n^3+2x_n<4$ $\endgroup$ Commented Sep 17, 2015 at 18:09
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$\begingroup$ 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. $\endgroup$– pjs36Commented Sep 17, 2015 at 18:19
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1$\begingroup$ @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 ? $\endgroup$ Commented Sep 17, 2015 at 18:23
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$\begingroup$ Yes, I was clearly confused there, you're exactly right. My apologies! $\endgroup$– pjs36Commented Sep 17, 2015 at 18:35
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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 $S$.
answered Sep 17, 2015 at 17:58
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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.
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$\begingroup$ Is it possible to say it isn't bounded below? $\endgroup$– LenolCommented Sep 17, 2015 at 18:05
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$\begingroup$ 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? $\endgroup$– LenolCommented Sep 17, 2015 at 18:16
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1$\begingroup$ @Achro, yes. Try it. If you get stuck, ask a separate question. $\endgroup$– lhfCommented Sep 17, 2015 at 18:17