Let $V$ be a subspace of $\mathbb R^3$ generated by the vectors $\{(1,1,1),(0,1,1)\}.$ Is the set $(\Bbb R^3 \setminus V) \cup \{(t,2t,2t):t \in \Bbb R \}$ connected in $\Bbb R^3 $
Attempt: V = Span$\{(1,1,1),(0,1,1)\}$ $= \{\lambda_1(1,1,1)+\lambda_2(0,1,1)~|~\lambda_1,\lambda_2 \in \Bbb R\}$ $=\{(\lambda_1,~(\lambda_1+\lambda_2),~(\lambda_1+\lambda_2)~|~\lambda_1,\lambda_2 \in \Bbb R\}$
Thus, elements in $V$ are of the form $\{(a,b,b)~|~a,b \in \Bbb R \}$
Let $Z= \{(t,2t,2t):t \in \Bbb R \} = \{t~(1,2,2):t \in \Bbb R \}$ .
Thus, $Z$ represents a line and $Z \subset V$.
Thus, $(\Bbb R^3 \setminus V) \bigcap Z= \phi$.
$Z$ is closed in $\Bbb R^3$ and thus, $(\Bbb R^3 \setminus V) \bigcap Z$ cannot be expressed as the union of two disjoint open sets. Thus, the set $(\Bbb R^3 \setminus V) \bigcup Z$ is connected in $\Bbb R^3$
Is this solution correct?
First, you remove a plane from $\mathbb{R}^3$, which clearly disconnects it. Then, you add a line on the removed plane, which connects your set again (you can go from one half-space to the other through the added line). Your set is even path-connected.