In the problem, I have to find the subset of the domain Z = {(x, y) $\in$ $\mathbb{R}^2$ | x, y > 0} for which the following function is continuous:
\begin{equation*}
h(x,y) = \left\{
\begin{array}{ll}
f(x,y) & \text{if x $\neq$ y} \\
g(x,y) & \text{if $x=y$}
\end{array}
\right.
\end{equation*}
I have proved that the $\lim_{(x,y) \to (n,n)} f(x,y)$ = g(n,n) when n =$\frac12$
I took this to mean that h(x,y) is continuous for {x>0, y>0, x $\neq$ y except when x=y=$\frac12$}
Could I write this as D = [f(x, y) $\in$ $\mathbb{R}^2$ | x>0, y>0, x $\neq$ y \ {x=y=$\frac12$}]?
I might have completely missed the mark, but any help is greatly appreciated, thanks!
I would write this as $$\{(x, y) \in (0, \infty) \times (0, \infty) \mid x \neq y\} \cup \left\{\left(\frac{1}{2}, \frac{1}{2}\right)\right\}.$$ Just as a word of caution, you shouldn't use the set difference when dealing with equations. Writing $x \neq y \setminus \{x = y = \frac{1}{2}\}$ doesn't make sense. I understand what you're going for (you're reading \ as "except for"), but this is not how the notation works. It's really only used for sets.