# Which of the following spaces are connected?

Let $$V$$ be the span of $$(1,1,1)$$ and $$(0,1,1) \in \Bbb R^3$$. Let $$u_1=(0,0,1),u_2=(1,1,0)$$ and $$u_3=(1,0,1)$$. Which of the following is/are correct?

$$1. \$$ $$(\Bbb R^3 \setminus V) \cup \{(0,0,0)\}$$ is not connected.

$$2. \$$ $$(\Bbb R^3 \setminus V) \cup \{tu_1+(1-t)u_3 : 0 \le t \le 1 \}$$ is connected.

$$3. \$$ $$(\Bbb R^3 \setminus V) \cup \{tu_1+(1-t)u_2 : 0 \le t \le 1 \}$$ is connected.

$$4. \$$ $$(\Bbb R^3 \setminus V) \cup \{(t,2t,2t):t \in \Bbb R \}$$ is connected.

$$(1),(3)$$ and $$(4)$$ are path-connected spaces so they are connected. Hence clearly $$(1)$$ is false and $$(3)$$ and $$(4)$$ are correct options. I think $$(2)$$ is also path-connected as $$(0,0,1) \in V$$. So the space is such that it is the three dimensional Euclidean space separated by the plane $$V$$ which is $$x=y$$ in such a way that the plane has a fracture along the line segment joining $$(1,0,1)$$ and $$(0,0,1)$$ . Now since $$(0,0,1) \in V$$ we can join any two points lying on two sides by joining them by a curve passing through $$(0,0,1)$$ not cutting any other point of the plane $$V$$. Is it not possible?

Your idea and logic are correct: If $V$ is a proper subspace and $S$ a subset of $\Bbb R^n$, then $(\Bbb R^n\setminus V)\cup S$ is connected and path-connected if and only if $S\cap V\ne \emptyset$. However, your result for 2. is false because the starting point is false: $(0,0,1)\notin V$. Note that $y=z$ for all points $(x,y,z)$ in $V$ and $y=0\ne 1=z$ for all points in the line segment.