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

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless I'm mistaken, this set is exactly the same as the interval \([0,1)\subset\mathbb{R}\), isn't it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I remember it right, a closed set is a set that contains its limits/accumulation points.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We use the complement set of the given set is open , hence the set is closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It's ok, thanks for being here. :)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Closed 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.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Let \(x\in S'\). For \(0\leq \operatorname{Re}(x) \leq1,\,\epsilon=\operatorname{Im}(x)\). For \(\operatorname{Re}(x)>1,\,\epsilon=x1\). For \(\operatorname{Re}(x)< 0,\,\epsilon=x\).
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