Open or closed sets in $\mathbb R^2$ Are the following two sets open or closed?
X = {$(x_1,0): x_1\in \mathbb  R$}   $\cup$   {$(0,x_2): x_2\in \mathbb R$} 
Y = {$(0,x_2): x_2\in \mathbb R$}
I'm really confused, before I though that both were open but I've been told that X is open and Y was closed. Can anyone explain why?
 A: If we're using the standard topology on $\mathbb{R}^2$, $X$ and $Y$ are both closed and non-open in $\mathbb{R}^2$ and $Y$ is thus closed in $X$ too. $Y$ is not open in $X$,as the origin is not an interior point.
A: $Y$ is the y-axis and $X$ is the union of the x-axis and the y-axis.
These are both closed sets in the plane. The easiest way to see that is to note that their complements are open sets: the complement of $Y$ is $$(U\times\mathbb R)\cup (V\times \mathbb R)$$ and the complement of $X$ is $$(U\times V)\cup (V\times V)\cup (U\times U)\cup (V \times U)$$ where $U$ and $V$ are the open intervals $(-\infty,0)$ and $(0,\infty)$ respectively. These are both unions of basic open sets, which are products of open sets.
Neither is an open set. This follows from the fact that they are both closed proper subsets and the whole space $\mathbb R\times\mathbb R$ is connected (so there are no nontrivial closed and open sets).
A: Assuming that we’re using the standard topology, it’s easy to show both sets are closed by showing their complements are open.  Let’s see this with $X$.  Choose an arbitrary point $(x, y) \notin X$.  If we can show that there’s a ball of positive radius around $(x, y)$ that does not intersect $X$, then we’re done.  Since $(x, y) \notin X, t= \min(|x|, |y|) \gt 0$.  Choose $\epsilon =t/2.$  Then $B((x, y), \epsilon) \cap X=\emptyset$.  A similar proof focusing only on the second coordinate works to show $Y$ is closed.
