I had seen answer given by member of MSE, of one of the question which was posted here earlier, here is link.
If $\mathbb R^3\setminus V$ connected where $V$ is the subspace generated by $\{(1,1,1),(0,1,1)\} $
But I didn't get, "how they concluded and how to prove"
$\left(\mathbb{R}^3\setminus V\right) \cup S$ will be connected if and only if $S\cap V\neq\emptyset$
Please anyone help me, I have same question as given in that link, but I am partially satisfied with that answer (as I am unable to prove above fact) though, I can skip the proof of above fact and directly get the answer using that fact, But then, it will not be learning, it will be just memorising and I am here to learn mathematics, not to memorize the mathematics! Please help me, to prove above statement(I know the definitions of connected and path-connected.)
It looks like the point he is making is that any if any point in $V$ is added to $\mathbb{R}^3$ \ $V$, then it will be connected.
In fact, it will be path connected since if we take any two points in the set, we can make continuous path between them with two segments, the first going straight from the first point to the point in $S \cap V$ and the second from there to the other point. In fact, this shows that it is polygonally connected.
We can turn this into a proof as follows:
Suppose $S \cap V \neq \emptyset$. Then, we can construct the path as described showing that it is path-connected an therefore connected. Specifically, Suppose $x, y \in (\mathbb{R}^3 $\ $V)\cup S$ and that $z \in V \cap S$ (whcih exist by hypothesis). The straight segment from $x$ to $z$ can be written as $f_1(t) = (1-t)x + tz$ for $0 \leq t \leq 1$. A similar path from $z$ to $y$ can be constructed (call this $f_2$).Then we can combine them into one path $\phi(t) = \begin{cases} f_1(2t),& 0 \leq t \leq \frac{1}{2}\\ f_2(2t-1),& \frac{1}{2} < t \leq 1\\ \end{cases} $
(one can verify that the path is contained in the set)
(This construction of line segments if good to know in general and the joining of paths to form $\phi$ can be used as part of showing that the ability of two points to reach each other by a path is an equivalence relation)
Suppose $S \cap V = \emptyset$. Then $(\mathbb{R}^3 $\ $V)\cup S = (\mathbb{R}^3 $\ $V)$ which is disconnected.
This completes the proof.
(edit: there is a subtle error in the proof: I will leave it as an exercise for the readers to find and fix it)