Loser66
  • Loser66
Is the set \(\{z in C : z ~is~ real~~ and ~~0\leq z <1\}\) closed or open? Please, help
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • 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
Loser66
  • Loser66
@SithsAndGiggles
anonymous
  • anonymous
Unless I'm mistaken, this set is exactly the same as the interval \([0,1)\subset\mathbb{R}\), isn't it?

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Loser66
  • Loser66
yes
anonymous
  • anonymous
If I remember it right, a closed set is a set that contains its limits/accumulation points.
Loser66
  • Loser66
We use the complement set of the given set is open , hence the set is closed
anonymous
  • 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
Loser66
  • Loser66
It's ok, thanks for being here. :)
thomas5267
  • 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.
thomas5267
  • 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 \(\operatorname{Re}(x)< 0,\,\epsilon=|x|\).

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