I am having the following question
- Let $\mathbb{R}^4$ be equipped with the standard topology induced by the Euclidean metric and let $f: \mathbb{R}^4 \rightarrow \mathbb{R}$ be the function given by $$ f\left(x_1, x_2, x_3, x_4\right)=x_1^2+x_2^2+x_3^2-x_4^2 $$ Put $A=f^{-1}(\{1\}), B=A \cap\left\{x \in \mathbb{R}^4: x_4=0\right\}, C=f^{-1}([0, \infty))$. Prove whether the following statements are true or false.
With the help of comments, I have the following attempt, please correct me if anything is wrong through my learning process, thanks again.
$A$ is not compact. A subset of $R^n$ is compact iff it is bounded and closed. $A$ is clearly not bounded as there exist no $r>0$ such that $d(a,b)<r$ for all $a,b$ in A, therefore A is not bounded and not compact.
$B$ is compact, as there is a constrain on $x_4$, $x_1^2+x_2^2+x_3^2=1$ which is bounded by $r=2\pi$, therefore $B$ is bounded. Also, $R^4-B$ is open as it can be covered by open balls so $B$ is closed. Hence, $B$ is compact
This is very wrong. By this same logic, you could also prove that $[0,1]$ is not compact, because it is a subset of $\mathbb R$ which is not compact. What's more, your logic actually implies that no compact subsets of $\mathbb R^4$ exist, which is absurd (since finite sets are always compact).
In fact, the property is "$A$ is compact if and only if it is a closed subspace of some compact space $X$", and this property cannot really be used to prove that a set is not compact. This is because the only way you could do that is to prove that everytime $A$ is a subspace of $X$, then either $A$ is not closed or $X$ is not compact, and that is equally hard to prove (since you would also have to cover the case when $X=A$, in which case $A$ is clearly closed, so you basically need to prove $A$ is not compact, to prove that $A$ is not compact.