Show $R=\{(x, y)$ : $|x|<1,|y|<1\}$ is an open set. 
Show that an open rectangular region $R=\{(x, y)$ : $|x|<1,|y|<1\}$ in $\mathbb{R}^{2}$ is an open set.

I assume the metric is the usual one. I do know that, "A subset $M$ of a metric space $X$ is said to be open if it contains a ball(open) about each of its points". But I couldn't manage to construction the balls. I guess $\left\{\frac{1}{n}\right\}$ sequence can help in this construction.
I was wondering: could changing the metric can also change the open characterization also?
Any help will be appreciated.
 A: Choose a point $(x,y)\in\mathbb{R}^2$ with $|x|,|y|<1$ and let $r$ be the shortest distance to the edge of the square. Then the circle centred at $(x,y)$ of radius $r$ is contained in the interior.
Another approach would be to note that $f,g:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y):=|x|$ and $g(x,y):=|y|$ are both continuous. Since the continuous preimage of an open set is open, we have that $$\{(x,y)\in\mathbb{R}^2:|x|,|y|<1\}=f^{-1}(-\infty,1)\cap g^{-1}(-\infty,1)$$ is the intersection of two open sets, and therefore open.
On your question about how the 'open characterization' depends on the metric: the collection of open sets is called a topology and does indeed depend on the metric. Two metrics that give the same open sets are said to be topologically equivalent. In the case of $\mathbb{R}^2$, the 'usual' metrics $$d_1((x_1,y_1),(x_2,y_2)):=|x_1-x_2|+|y_1-y_2|,$$ $$d_2((x_1,y_1),(x_2,y_2)):=(|x_1-x_2|^2+|y_1-y_2|^2)^{\frac{1}{2}},$$ and $$d_\infty((x_1,y_1),(x_2,y_2)):=\text{max}\{|x_1-x_2|,|y_1-y_2|\}$$ all have the same open sets. You can see this by checking that there exists $c,C>0$ for which $cd_1(x,y)\leq d_2(x,y)\leq Cd_1(x,y)$ for each pair of points $x,y\in\mathbb{R}^2$ and similarly for comparing the other metrics. The collection of open sets that arises from these metrics is called the Euclidean topology, which is the one we are all familiar with.
A: Let $\epsilon = \min(1-|x|, 1-|y|)$ and consider the open ball about $(x,y)$ of radius $\epsilon$.
Need to show that this is contained in $R$.
Consider the (open) square region $Q$ whose sides are of length $2\epsilon$ that is centered at $(x,y)$.  This square region is completely contained in $R$.
Further, the open ball of radius $\epsilon$, centered at $(x,y)$ is completely contained in $Q$.
A: Take $(x,y) \in R$ and let $\delta = \operatorname{min}(1-|x|,1 -|y|)$. Then $B_{\delta}(x,y) \subset R$.
If you have a metric you define the topology in the most cases to be induced by the metric, i.e. open metric balls are open. Openness is a question of topology and the topology can be defined independently of any metric. If you change the metric the topology changes also in general. Take for example the metric $d(x,y) = 1$ if $x \neq y$ and $d(x,y) = 0$ if $x = y$. Then the $\frac{1}{2}$-balls are points, and therefore the topology is discrete.
A: Notice that :
$$R = ]-1, 1[ \times ]-1, 1[$$
then $R$ is open.
