why set $S$ is not closed under addition? some  doubts :
Consider $\mathbb{R}^2$  over $ \mathbb{R}. $
Then 
Given  set  $ S = \{(x,0) : x \in \mathbb{R}\} \cup \{ (0,y) : y \in \mathbb{R}\}$
My question  is that   why  set $S$ is not closed under addition?
My attempt  :  I think  set $S$  is  closed under the addition because  $$ S = \{(x,0) : x \in \mathbb{R}\} \cup \{ (0,y) : y \in \mathbb{R}\}= \{(x,y) : x,y \in \mathbb{R}\}$$
Any hints/solution
thanks u

First of all, do you agree that $(1,1) \not\in \{(x,0) : x \in \mathbb{R}\}$ and $(1,1) \not\in \{(0,y) : y \in \mathbb{R}\}$? If yes, then it is obviously not in their union.
You are confusing two different concepts, I think.
Consider the plane ($\mathbb{R}^2$) and then let $l_1$ be the line $\{(x,y)\in\mathbb{R}^2: y=x\}$ and let $l_2$ be the line $\{(x,y)\in\mathbb{R}^2: y=-x\}$.
Obviously, the union of these two lines is not a vector space because $(0,1)$ is not on any of these lines. But the space generated by these two lines is all of the plane. Can you see why?
This is similar to your case: $$S = \underbrace{\{(x,0) : x \in \mathbb{R}\}}_{S_1} \cup \underbrace{\{ (0,y) : y \in \mathbb{R}\}}_{S_2} \subsetneq \{(x,y) : x,y \in \mathbb{R}\}$$ because $(1,0) \in S_1$ and $(0,1) \in S_2$ but $(1,1)$ is neither in $S_1$ nor in $S_2$. So, it can't be in $S=S_1 \cup S_2$ because by definition, if something is in $S_1 \cup S_1$, then it belongs to $S_1$ or $S_2$.
However, it is true that $\langle S_1 \cup S_2\rangle=\mathbb{R}^2$. Can you see why?
