Define the vector space $V = \{ (a,0,0,b) : a,b \in \mathbb{R} \} \subseteq \mathbb{R^4}$.
Is $W = \{ (a,b,c,d) \in \mathbb{R^4} : ab = c \}$ a subspace of $\mathbb{R^4}$ ?
Is $W = \{ (a,b,c,d) \in V : ab = c \}$ a subspace of $V$ ?
To be a subspace of $\mathbb{R^4}$, it needs to have the zero vector. For $\mathbb{R^4}$, we have the zero vector $(0,0,0,0)$. And we have $0 \cdot 0 = 0$, so it is included in the set.
We also need to check that it is closed under addition and scalar multiplication. This is not the case because in general $(a_1 + a_2) \cdot (b_1 + b_2) \neq c_1 + c_2$. So it is not closed under addition, thus it is not a subspace.
To be a subspace of $V$, we have $a \cdot 0 = 0$ which is always true for any $a$. So it contains the zero vector.
Now, we need to check that it is closed under addition and scalar multiplication. The condition $ab =c$ will always hold because $b = c = 0$ given how $V$ is defined.
So $W$ would be a subspace of $V$, but not a subspace of $\mathbb{R^4}$. Is that correct ?
You are correct. The set $W$ is not a subspace of $\mathbb{R}^4$.
Note that $(1,2,2,d)\in W$, because $1\cdot2=2$ and $(1,1,1,d)\in W$, because $1\cdot1=1$. However, $(1,2,2,d)+(1,1,1,d)=(2,3,3,2d)$ is not in $W$, because $2\cdot3\neq3$. Therefore, the set $W$ is not a subspace of $\mathbb{R}^4$.
You are also correct in claming that $W$ is a subspace of $V$. Note that if $(a,b,c,d)\in V$, then $b=c=0$. Therefore, $W=\{(a,b,c,d)\in V:ab=c\}=\{(a,0,0,d):a,d\in\mathbb{R} \ \ \ \text{and} \ \ \ a\cdot0=0\}$. Since any real number $a$ satisfies $a\cdot0=0$, then $W=\{(a,0,0,d):a,d\in\mathbb{R} \ \ \ \text{and} \ \ \ a\cdot0=0\}=\{(a,0,0,d):a,d\in\mathbb{R}\}=V$. Therefore, $W$ is a subspace of $V$, because $W=V$.