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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

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  1. Loser66
    • one year ago
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    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
    • one year ago
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    @SithsAndGiggles

  3. anonymous
    • one year ago
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    Unless I'm mistaken, this set is exactly the same as the interval \([0,1)\subset\mathbb{R}\), isn't it?

  4. Loser66
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    If I remember it right, a closed set is a set that contains its limits/accumulation points.

  6. Loser66
    • one year ago
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    We use the complement set of the given set is open , hence the set is closed

  7. anonymous
    • one year ago
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    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
    • one year ago
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    It's ok, thanks for being here. :)

  9. thomas5267
    • one year ago
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    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
    • one year ago
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    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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