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manjuthottam

  • 3 years ago

Theoreticals of Calculus: could someone explain indexed family of sets to me??

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  1. manjuthottam
    • 3 years ago
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    Here is a pdf of a question and solution involving indexed family sets on page 5 & 6! please and thankyou!!

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  2. manjuthottam
    • 3 years ago
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    is there a difference between {} and [] ? Also I am making more mistake on finding the intersection portion of the set

  3. KingGeorge
    • 3 years ago
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    Well, whenever you see { and }, those denote generic sets. In this case, [ and ] are denoting an interval on the real number line. So [1,2] is just every real number \(x\) such that \(1\le x\le2\). So if you have \[\left\{\left[1,1+\frac{1}{n}\right]:n\in\mathbb{N}\right\},\] this is the set of all intervals \([1,1+1/n]\) such that \(n\) is a natural number greater than 0.

  4. KingGeorge
    • 3 years ago
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    And can you be a little more specific on what parts about the intersection are confusing you?

  5. manjuthottam
    • 3 years ago
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    oh ok i understand now about the {} and[]! about the intersection, on page 5 the intersection is {1} but why on page 7 the intersection is empty set?

  6. manjuthottam
    • 3 years ago
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    i thought the answer would be 1 as well

  7. KingGeorge
    • 3 years ago
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    Ah. On page 7, they use ( and ) instead of [ and ]. The difference between these two, is whether the endpoints are included. So if you have [1,2], this is every real number x such that \(1\le x\le2\). However, (1,2) is every real number x such that \(1<x<2\).

  8. KingGeorge
    • 3 years ago
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    You can also mix and match. So (1,2] would be all real numbers x such that \(1<x\le2\). All in all, on page 7, you have an empty intersection because 1 is never included in any of the sets.

  9. manjuthottam
    • 3 years ago
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    oh so since < > are used, 1 is not part of the set, just the smallest number slightly greater than it is?

  10. KingGeorge
    • 3 years ago
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    Precisely.

  11. manjuthottam
    • 3 years ago
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    that makes sense thank you!!

  12. KingGeorge
    • 3 years ago
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    You're welcome.

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