Is the set $(\Bbb R^3 \setminus V) \cup \{(t,2t,2t):t \in \Bbb R \}$ connected in $\Bbb R^3?$

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?

• The set would even be connected if $Z$ was a set containing only one element of $V$. Apr 30, 2020 at 9:47

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.

• Thank you. Interesting Answer. Do you think my solution is correct as well? Apr 30, 2020 at 9:07
• Especially the part : $Z$ 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 $(\Bbb R^3 \setminus V) \bigcap Z$ is connected in $\Bbb R^3$ Apr 30, 2020 at 9:08
• I do not quite understand your argument. How do you derive from the fact that your set is the union of disjoint sets, one of them open and the other one closed, that it is impossible to represent it as a union of disjoint non-empty open sets? Apr 30, 2020 at 9:15