# Thread: closed and bounded sets

1. ## closed and bounded sets

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.

{(x^3y - z^2; xyz + y^3z) : 0 =< x, y,z =< 1).}

I finding it difficult to show this is closed.

Thanks for helping

2. 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
.

3. Originally Posted by tonio
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How does one show the second set is closed since it has two functions.

The first one is closed since the complment being open means set is closed

Thanks for helping me