Is this set open or closed (or both?) I'm trying to figure out whether or not the following set is open or closed.
$$D=\{(x,y,z)\in\mathbb R^3\mid x\gt0,y\gt0,z=0\}$$
I've tried imagining it and to me, it seems like an open set, but maybe it is both open and closed. How would I determine that?
 A: No, it is not an open set. For instance, $(1,1,0)\in D$, but no open ball centered at $(1,1,0)$ is contained in $D$.
On the other hand, $\left(\left(\frac1n,\frac1n,0\right)\right)_{n\in\mathbb N}$ is a sequence of elements of $D$ which converges to $(0,0,0)$. But $(0,0,0)$ does not belong to $D$. What can you deduce from this?
A: It is not open because it contains $(1,1,0)$ and every neighborhood if this point contains points with $z \neq 0$. It is not closed because $(\frac  1 n, \frac  1 n,0)$ is a sequence in this set which converges to a point outside the set.
A: Theorem: Let x$\in \mathbb{R^n}$  where the topology induced by the standard metric is assumed. $x\rightarrow a $ $\iff$ $x^i \rightarrow a^i$ for each $1\leq i\leq n$
Take the sequence $(\frac{1}{n},\frac{1}{n}, 0)$ 
Theorem 2: Let X be a metric space. $A \subset X$ is closed  iff the limit of every convergent sequence in A is in A.
Observe that the components of the above sequence  all converge to 0 and therefore the sequence converges to $(0,0,0)$ which is not in the set.  To show that the set is open, you must find a point in the set such that no matter what radius your ball has, it is never contained in the set. The previous answers wrote $(1,1,0)$ you should prove that what they wrote indeed shows that the set is not open.
