$S_{1}=\{ (x_{1},x_{2}):x_{1}^2+x_{2}^2=1 \}$
I cant solve trying with an alpha value $\in[0,1]$ by two dummy vectors. Please can anyone tell me how to prove it is not convex set ?
what I tried is
Defined two vectors $$a=(a_{1},a_{2})\in S_{1}$$ so $$a_{1}^2+a_{2}^2=1$$, $$b=(b_{1},b_{2})\in S_{1}$$ so $$b_{1}^2+b_{2}^2=1$$,
and tried to prove $$\alpha\begin{bmatrix} a_{1}\\a_{2} \end{bmatrix}+(1-\alpha)\begin{bmatrix} b_{1}\\b_{2} \end{bmatrix}\not\in S_{1} ? ?$$
it was so hard to me continuing this
To show that a set $S$ is convex, you must prove that
"For all $x,y \in S$ and $\alpha \in [0,1]$, we have $\alpha x+(1-\alpha)y \in S$."
To show that a set $S$ is not convex, you need to prove the negation of the above statement, i.e.
"There exists $x,y \in S$ and $\alpha \in [0,1]$ such that $\alpha x+(1-\alpha)y \not\in S$."
In your particular problem, this means that you don't have to prove $\alpha x+(1-\alpha)y \not\in S$ for every choice of $x,y \in S$ and $\alpha \in [0,1]$, but rather just one particular choice.
So in your problem, pick any two distinct points in $S$ (say $x = (1,0)$ and $y = (-1,0)$) and some value of $\alpha \in [0,1]$ (say $\alpha = 1/2$), and see if you can check that $\alpha x+(1-\alpha)y \not\in S$ for this choice of $x$, $y$, and $\alpha$.