Let $$S=\{x\in \mathbb{R} : \text{ } x^6-x^5 \leq 100\}$$
$$T=\{x^2-2x : \text{ } x\in (0, \infty)\}$$
Determine wether $ S \cap T$ is closed and bounded?
My Try
$S$ is bounded hence $ S \cap T$ is also bounded.
$S=[a,b]$ where $a<-1$ because -1 satisfy the equation given. and $T=(-1, \infty)$ hence $ S \cap T=[-1,b]$
Hence $ S \cap T$ is closed and bounded
However I am not satisfied with my reasoning. Can anyone tell me the proper solution.
Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-\infty, 100])$ is closed.
Since $s(x) \to \infty$ as $|x| \to \infty$ we see that $S$ is bounded, hence compact.
Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $\min$ of $-1$. at $x=1$, and we see that $t(x) \to \infty$ as $x \to \infty$. Hence $T=[-1,\infty)$, which is closed.
Hence $T \cap S$ is closed.