# If $\mathbb R^3\setminus V$ connected where $V$ is the subspace generated by $\{(1,1,1),(0,1,1)\}$

Let $V$ be a subspace of $\mathbb R^3$ generated by the vectors $\{(1,1,1),(0,1,1)\}.$ There are $3$ vectors $u_1=(0,0,1),u_2=(1,1,0),u_3=(1,0,1)$.Then which of the following sets are connected?

$1.\left(\mathbb R^3\setminus V\right)\cup\{(0,0,0)\}$

$2.\left(\mathbb R^3\setminus V\right)\cup \{tu_1+(1-t)u_3:t\in [0,1]\}$

$3.\left(\mathbb R^3\setminus V\right)\cup\{tu_1+(1-t)u_2:t\in [0,1]\}.$

$4.\left(\mathbb R^3\setminus V\right)\cup \{(t,2t,2t):t\in \mathbb R\}$

Now $V$ is generated by two linearly independent vectors so is like $\mathbb R^2.$ So $\left(\mathbb R^3\setminus V\right)$ will be disconnected but giving the $(0,0,0)$ with them will make it connected.

No. $4$, I think will be connected also because $\{(t,2t,2t),t\in \mathbb R\}$ is in $V$ and contains the point $(0,0,0)$ Now, any neighbourhood of this point will have points from both $V$ and $\left(\mathbb R^3\setminus V\right)$ and so the union is connected.I'm not sure about no. $2$ and $3.$

I believe it's good to think about this problem geoemetrically. $V$ is a plane in $\mathbb{R}^3$, to so $\left(\mathbb{R}^3\setminus V\right) \cup S$ will be connected if and only if $S\cap V\neq\emptyset$. Can you see why?
With this in mind, the problem boils down to figuring out which of the sets $S$ intersect $V$ in each of the four cases.
• Sorry but I am unable to prove the fact that $(\Bbb R^3\setminus V)\cup S$ is connected $\iff S\cap V\neq \emptyset$ ...Did you prove that? Commented Dec 5, 2016 at 8:10
• Okay,let's start; suppose that $S\cap V=\emptyset$ How to claim that $(\Bbb R^3\setminus V) \cup S$ is connected Commented Dec 5, 2016 at 8:17
• You misunderstand; if $S\cap V = \emptyset$, then $\left(\mathbb{R}^3\setminus V\right) \cup S$ is disconnected. Notice that, in this case, $S\subset \left(\mathbb{R}^3\setminus V\right)$ and hence $\left(\mathbb{R}^3\setminus V\right) \cup S = \left(\mathbb{R}^3\setminus V\right)$. Commented Dec 5, 2016 at 8:23
• For case $3$, take $a=t=1/2$ and $b=0$. Commented Dec 7, 2016 at 6:12