Is A open or closed? Is it a bounded set? Verify if the given set $A$ it's open or closed. Also verify if it's bounded.
a) $A = \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 < 1\}$
I just want the method for verify this things in case of this example, because I have items a,b,c,d,e to do. The other sets are usually points of $\mathbb{R}^2$.
 A: Some highlights: suppose $\;\{(x_n,y_n)\}\subset \Bbb R^2\setminus A\;$ convergent, then we have that
$$\begin{cases}x_n\xrightarrow[n\to\infty]{}x_0\\{}\\y_n\xrightarrow[n\to\infty]{}y_0\end{cases}\;\;\;\text{and}\;\;(x_0,y_0)\in\Bbb R^2\setminus A$$
as we know from limits in the real line, and thus the complement of $\;A\;$ is closed so $\;A\;$ is open. It also is trivially bounded, say by the unit circle.
A: Intuition should tell you this is the interior but not the edge of a unit circle and is therefore open, not closed and bounded.
But intuition is not math.
To show something is not closed, show a limit point is not in the set.  Points on the circle are limit points.  So take any $(x,y)$ where $x^2 + y^2 = 1$ and show it is a limit point.  The easiest would be $(x,y) = (1,0)$.  Let $N_{\epsilon}(x,y)$ be a neighborhood.  $(1 - \epsilon/2,0)$ is in the neighborhood.  $(1 - \epsilon/2)^2 + 0^2 < 1$.  So $(1,0)$ is a limit point.  $(1,0)$ is not in the set.  So it is not closed.
Showing it's open is a bit of a pain but show if $(a,b) \in A$ so $a^2 + b^2 < 1$ Let $\epsilon = \sqrt{1 - (a^2 + b^2)}$.  Show $N_{\epsilon/2}(a,b) \subset A$ via triangle inequality.  
Show $A$ is bounded via triangle inequality.  $d((a,b),(c,d)) \le d((a,b)(0,0)) + d((c,d),(0,0)) = \sqrt {a^2 + b^2} + \sqrt{c^2 + d^2} < 1 + 1 = 2$.  So $A$ is bounded.
A: we can paremeterize it $x=rcos\theta ,y=rsin\theta $
$A = \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 < 1\}$
then we can Write $r^2\lt1 $
no matter how small $\epsilon $ ball you take in $A$ it will always be contained in $A$ hence the Set A is an open set.
