$\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ open or closed 
Possible Duplicate:
Proving that a complex set in open/closed/neither and bounded/not bounded 

I think $\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ is closed. But I have no idea how to show it since you have to take an element of the set (which lies on the axes) and take a neighbourhood around that (not all of the ball is within the set so it's not open). 
But when you do the complement, not all of everything outside of the axes includes the ball (as some of the ball is on the axes). This would make it neither open nor closed but I'm sure it's closed! Can someone help me please? 
It's definitely not bounded because no closed ball can cover all of the axes as they go on to infinity and beyond. Right?

If you mean the set $\{z\in\mathbb C : |z|=|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\}$ you might want to consider the continuous function $g(z)=|z|-|\mathrm{Re}(z)|-|\mathrm{Im}(z)|$ and remember how continuous functions behave with closed sets.
