I am trying to solve the following problem (not homework):
Equip $\mathbb R^4$ with the standard topology induced with the Euclidean metric and let $f:\mathbb R^4 \rightarrow \mathbb R$ be the function given by $$f((x_1,x_2,x_3,x_4) = x_1^2 - x_2^2 + x_3^2 - x_4^2.$$ Define the sets $A = f^{-1}(\{1\}), B = A \cap \{x \in \mathbb R^4 : x_2^2 + x_4^2 = 1\}$ and $C = f^{-1}((0,\infty)) \cap \{x \in \mathbb R^4: x_2=x_3=x_4=0 \}$.
I need to determine whether the following statements are true or false.
a) $A$ is compact,
b) $B$ is compact,
c) $B$ is path-connected,
d) $C$ is connected
I have managed to get an answer for a), b) and d), although I am not sure if my reasoning is correct. On c) I got completely stuck, so any hints or answers would be appreciated. Here is my attempt:
a) The singleton $\{1\}$ is closed under the standard topology on $\mathbb R$ $\Rightarrow A = f^{-1}(\{1\}) $ is closed in the standard topology on $\mathbb R^4$ if $f$ is continuous. Since $f$ is differentiable, it is also continuous. However, $A$ is not bounded (there is no upper limit to the distance between two points that satisfy the equation $x_1^2 - x_2^2 + x_3^2 - x_4^2 = 1 \iff x_1^2 + x_3^2 = 1 + x_2^2 + x_4^2 $). Heine-Borel states that for $K \subseteq \mathbb R^n, \mathbb R^n$ equipped with the standard topology we have: $K$ compact $\iff K$ closed and bounded. Hence, A cannot be compact.
b) $A$ is the set of points $x=(x_1,x_2,x_3,x_4)$ for which $x_1^2 - x_2^2 + x_3^2 - x_4^2 = 1$. To get $B$, we only want to keep the points where $x_2^2 + x_4^2 = 1$, hence $B = \{x \in \mathbb R^4: x_1^2+x_3^2=0, x_2^2 + x_4^2 = 1 \}$, or more simply $B = \{x \in \mathbb R^4: x_1 = x_3 = 0, x_2^2 + x_4^2 = 1 \}$. We can say that $B$ is closed (since its compliment is open). It is also bounded (it is a cirlce in $\mathbb R^4$ space). By Heine-Borel's theorem I then conclude that $B$ is compact.
c) I have no idea where to begin here. I was initially thinking of the theorem: An open set in $\mathbb R^n$ is connected if and only if it is path-connected. However, to me $B = \{x \in \mathbb R^4: x_1 = x_3 = 0, x_2^2 + x_4^2 = 1 \}$ seems to be closed by the same reasoning as above. Also, $B$ seems to be connected so I can't use that to disprove that it is not path-connected.
d) $C$ is the set of points for which $x_1^2 - x_2^2 + x_3^2 - x_4^2 > 0$ and $x_2=x_3=x_4=0 \iff x_1^2 > 0, x_2=x_3=x_4=0 \iff x_1 \neq 0, x_2=x_3=x_4=0$, that is $C=(-\infty,\infty)\setminus 0 \times \{0\}^3$. A possible seperation for this set would be $U = \bigcup_{i=1}^{\infty} B_{r=1}^d(x=(i,0,0,0))$ and $U = \bigcup_{i=1}^{\infty} B_{r=1}^d(x=(-i,0,0,0))$ where $d$ is the Euclidean metric on $\mathbb R^4$. Now $C \subseteq U \cup V, U \cap V = \varnothing, U \cap C \neq \varnothing \neq V \cap C$. Hence $C$ is not connected.
Please let me know if any clarification is needed, thanks.
We have $A=\{(x_1,x_2,x_3, x_4)\ |\ x_1^2-x_2^2+x_3^2-x_4^2=1\}$. Then $B$ is defined by the following system of equations
$$\begin{cases} x_1^2-x_2^2+x_3^2-x_4^2=1 \\ x_2^2+x_4^2=1 \end{cases}$$
which is equivalent to
$$\begin{cases} x_1^2+x_3^2=2 \\ x_2^2+x_4^2=1 \end{cases}$$
or in other words $B$ is homeomorphic to product of two spheres $S^1\times S^1$ via the following map:
$$S^1\times S^1\to B$$ $$\big((x_1,x_2),(x_3,x_4)\big)\mapsto \big(\sqrt{2}\cdot x_1, x_3,\sqrt{2}\cdot x_2, x_4\big)$$
I leave as an exercise that this is a homeomorphism. This should answer both path-connectedness and compactness questions.
Your other reasonings seem to be fine.