Originally Posted by
nerdo working in (Real)^2
{(x, 0) : sin x ≥ 0}
I Know this set is closed but is unbouned. But i am finding it different to show this.
Did you mean "difficult" instead of "different"? Let $\displaystyle \displaystyle{X:=\{(x,0)\in\mathbb{R}^2\;;\;\sin x\geq 0\}=\bigcup\limits_{n\in\mathbb{Z}}[2n\pi,\,(2n+1)\pi]}$ .
Now, let $\displaystyle x_0\notin X\Longrightarrow \sin x_0<0\Longrightarrow \exists\epsilon > 0\,s.t.\,\,\sin x<0\,\,\forall\,x\in(x_0-\epsilon,\,x_0+\epsilon)$ (why??) , and
thus the complement of $\displaystyle X$ is an open set (why?)
{(x^3y - z^2; xyz + y^3z) : 0 =< x, y,z =< 1).}
Try to do something similar to the above
Tonio
I finding it difficult to show this is closed.
Thanks for helping