How do you show without a graph that the first set $T$ is not compact and that the second set $C$ is compact? So I'm a little stuck here.
Initially, the question was to determine if either of these sets were compact, and I had concluded that the first was and that the second wasn't.
I then went to check Desmos to verify my answers and found quite the opposite to be true, so I'm curious about how to prove this analytically?
I tried again but I'm entirely lost. Here were the sets:
Let $T$ be a subset of $R^2$ such that, $T = \{ (x,y) \in R^2 : x^3 + y^3 + xy \leq 25\}$
Let $C$ be a subset of $R^2$ such that, $C = \{ (x,y) \in R^2 : x^4 + y^{18} \leq 25 \}$
Thanks in advance, any advice or direction would be awesome. I'm aware of the Heine–Borel theorem but can't figure out how to prove we can use it for the set $C$
 A: Both sets are closed and therefore each of them is compact if and only if it is bounded.
The set $T$ is unbounded because, for each $y\in\Bbb R$, the equation (in the variable $x$) $x^3+y^2+xy=0$ has some solution $x$ (it's a cubic equation) and therefore $T$ has points whose second component is arbitrarily large. Therefore, $T$ is unbounded.
But if $(x,y)\in\Bbb R^2$ is such that $|x|,|y|\geqslant3$, then $x^4+y^{18}>25$ and therefore $(x,y)$ doesn't belong to $S$. Therefore, $C$ is bounded.
A: $T$ is not bounded. You can calculate and see that $(-n, 0) \in T$ for any $n \in \mathbb N$. So $T$ compact is impossible.
$C$ is bounded. Indeed, for a crude bound, $C$ is contained in the disk $D(0; 5) = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \leq 25\}$. To see $C$ is closed, note that the function $f_C(x, y) = x^4 + y^{18}$, $\mathbb{R}^2 \to \mathbb{R}$ is continuous and $C$ is the preimage of the closed set $[0, 25]$ in $\mathbb{R}$. Hence, $C$ is closed also. So by Heine-Borel, compactness follows.
A: Hint: A subset of $\mathbb{R}^2$ is compact iff it is both closed and bounded. Since each is defined by a single inclusive inequality, each is closed. Now it only remains to decide which is bounded, I leave that to you.
