Showing complement of subsets are open How do you show the complement of these subsets are open?


*

*$S^1=\{(x_1,x_2)\in\Bbb R^2\mid x^2_1+x^2_2=1\}$

*$B^c_1=\{(x_1,x_2)\in\Bbb R^2\mid x^2_1+x^2_2\ge1\}$

 A: I'll do the second problem:
$\color{darkgreen}{B_1}$ is the unit ball of radius 1 centered at the origin.  Note that it contains its boundary.  Any point at distance 1 from the origin is on $\color{darkgreen}{B_1}$.
Now, to prove that $B_1^C$ is open, you have to show the following:
for any point $\color{maroon}{x_0}$ that is not in $\color{darkgreen}{B_1}$, there is some  open ball $\color{maroon}{B_{x_0}(\epsilon)}$, centered at $\color{maroon}{x_0}$, that is disjoint from $\color{darkgreen}{B_1}$.
Here, $\color{maroon}{B_{x_0}(\epsilon)} =\{\,z\in\Bbb R^2 \mid d(x_0,z)<\epsilon\,\}$.
So, first pick a point $\color{maroon}{x_0}$ that is not in $\color{darkgreen}{B_1}$.  We know then that the distance from $\color{maroon}{x_0}$ to the origin is $r>1$ for some real number $r$.
We need to find an open ball centered at $\color{maroon}{x_0}$ that is disjoint from $\color{darkgreen}{B_1}$.  Note that the distance from $\color{maroon}{x_0}$ to the boundary of $\color{darkgreen}{B_1}$ is $r-1>0$ (recall $r>1$). We suspect that, if we let the radius, $\epsilon$, of our ball be less than $r-1$, then the open ball $\color{maroon}{B_{x_0}(\epsilon)}$ will "work".  
So, let's take the open  ball to be $B_{x_0}({r-1\over2})$.
Now,  we have to prove that this open ball works:
Of course, by definition
$B_{x_0}({r-1\over2})$ is centered at  $\color{maroon}{x_0}$.
Now we have to show that $B_{x_0}({r-1\over2})$  is disjoint from $\color{darkgreen}{B_1}$. To do this, we have to show that if $y\in B_{x_0}({r-1\over2})$, then $y\notin \color{darkgreen}{B_1}$. 
So let $y\in B_{x_0}({r-1\over2})$. We have to show that the distance from $y$ to the origin is greater than $1$. 
Towards this end, using the reverse triangle inequality, and the fact that $y$ being in 
$B_{x_0}({r-1\over2})$ guarantees that $d(y,x_0)<{r-1\over 2}$:
$$
d(y,0)\ge\bigl|\, d(x_0,0)-d(y,x_0)\,\Bigr|> r- {r-1\over 2} ={r+1\over2}>1.
$$
And we are done. Any point outside of $B_1$ is contained in an open ball disjoint from $B_1$; thus $B_1^C$ is open.

