Showing a set is open but not closed Let $X = \{ (x,y) \in \mathbb{R}^2 : x > 0, y \geq 0 \} $. I am claiming $X$ is open, but it is not closed. 
My Try:
To show it is not closed, I found a sequence that converges to a point outside $X$. For instance, $(x_n, y_n) = (\frac{1}{n}, \frac{1}{n} ) \to (0,0) \notin X$. Hence, $X$ cannot be closed. 
In trying to show it is open, I have diffictulty trying to find a suitable open ball contained in $X$. Any help would be greatly appreaciated. 
Thanks.

Draw a picture; this set is the first quadrant in the plane, together with the ray $$\{(x, y) : y = 0, x \ge 0\}$$
Any ball whose center has $y$-coordinate zero will intersect the lower half-plane, and so cannot be contained in the set; hence it's not open.

Morally, the set contains a little bit of its boundary, which causes it to be non-open.
