Showing a set is a compact subset of $\mathbb{R}$ Question: Let $$A=\{ x \in \mathbb{R}: x(x^{3}-3x-1)\leq15 \}.$$    
Show that A is a compact subset of $\mathbb{R}$.
I am just wondering how to approach this problem. Should I try showing it directly? Or is it enough to show that the set is closed and bounded in $\mathbb{R}$? Thanks
 A: This is a cool question and it's probably best to do it using sequences.
First note that the quartic $q(x)=x(x^3-3x-1)-15=x^4+\ldots\,$ goes to $+\infty$ as $x\longrightarrow\pm\infty$, so the set $A$ of all $x\in\mathbb{R}$ with $q(x)\leq 0$ is bounded - this can be easily made rigorous. Instead, you could also use of the fact that $x^3-3x-1=1+(x-2)(x+1)^2$ is $\geq 1$ if and only if $x\geq 2$.
For closedness, I'd use the sequential characterization: Let $(x_n)\subseteq A$ converge to some $x\in\mathbb{R}$. Then $x_n(x_n^3-3x_n-1)$ converges to $x(x^3-3x-1)$ by standard limit theorems. Since $x_n(x_n^3-3x_n-1)\leq 15$ we also get $x(x^3-3x-1)\leq 15$ by an elementary theorem. So $x\in A$.
If you are not comfortable using Heine-Borel here, you could use the fact that $A$ being compact is equivalent to every sequence in $A$ having a convergent subsequence in $A$: Starting with a sequence in $A$, you know it's bounded (as above), so Bolzano-Weierstrass gives you a convergent subsequence $\longrightarrow x\in\mathbb{R}$. Then you show that $x\in A$ by essentially the above closedness argument.
A: I guess a more general solution can be given as follows: Let $p$ be any polynomial, and $a$ be a constant real number. Then, since as $x$ goes to $\infty$ or $-\infty$ we know polynomial won't be bounded, we conclude that the set of $x$ satisfying $p(x)\leq a$ must be bounded. On the other hand, consider the set of $x$ such that $p(x)>a$. For each $x$ from this set, by the continuity of $p$, there will be an open neigborhood of $x$ that $p(x)>a$ remains true, so complement of our  original set is open, so our set is closed. Hence, our set is always compact.
