## Loser66 one year ago Is the set $$\{z in C : z ~is~ real~~ and ~~0\leq z <1\}$$ closed or open? Please, help

1. Loser66

I know it is not closed nor open but I am not confident on the proof For not open: let z =0, $$B(0,\varepsilon)$$ contains points $$\notin$$ the set, hence it is not open

2. Loser66

@SithsAndGiggles

3. anonymous

Unless I'm mistaken, this set is exactly the same as the interval $$[0,1)\subset\mathbb{R}$$, isn't it?

4. Loser66

yes

5. anonymous

If I remember it right, a closed set is a set that contains its limits/accumulation points.

6. Loser66

We use the complement set of the given set is open , hence the set is closed

7. anonymous

Well, it's been much longer than I thought... I'm afraid I won't be able to help much unless I spend quite a bit of time reviewing. A few others might be able to assist you. @ganeshie8 @thomas5267

8. Loser66

It's ok, thanks for being here. :)

9. thomas5267

Closed I guess. Let $$S=\{z:z \in \mathbb{R} \land 0\leq z <1\}$$. Let $$\epsilon$$ be the minimum distance from a $$x$$ in $$S'$$ and $$S$$. Such $$\epsilon$$ should work.

10. thomas5267

Let $$x\in S'$$. For $$0\leq \operatorname{Re}(x) \leq1,\,\epsilon=\operatorname{Im}(x)$$. For $$\operatorname{Re}(x)>1,\,\epsilon=|x-1|$$. For $$\operatorname{Re}(x)< 0,\,\epsilon=|x|$$.