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anonymous

  • 4 years ago

"Every finite set is closed." Is this true?

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  1. anonymous
    • 4 years ago
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    Yes, because for a finite value, there is a definite number which is the last one, so you can close it.

  2. anonymous
    • 4 years ago
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    Yes it is

  3. anonymous
    • 4 years ago
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    This is what I have: Proof: Let \(A\) be a finite set. Then \(A=\{x_1,x_2,...,x_n\}\) for some \(n\in\mathbb{N}\). Since every point in \(A\) is an isolated point. Therefore, \(A'=\varnothing\), and since \(\varnothing\in A\), it follows that \(A\) contains all its accumulation points. Therefore, \(A\) must be closed. \(\blacksquare\)

  4. JamesJ
    • 4 years ago
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    You mean in R^n, under the usual topology? Yes. Now, in your proof, there's something odd going on here. What is A', the compliment of A?

  5. anonymous
    • 4 years ago
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    It's the set of all its accumulation points.

  6. anonymous
    • 4 years ago
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    (or limit points)

  7. anonymous
    • 4 years ago
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    I have a theorem that states that if a set contains all its accumulation points, then it must be closed.

  8. JamesJ
    • 4 years ago
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    Ok, and then you mean to write \( \emptyset \subset A \). This is slightly obscure proof; I would use the definition of open and closed directly, as opposed to relying on such a result that comes farther down the road.

  9. anonymous
    • 4 years ago
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    Another hint for a direct proof: any FINITE union of closed sets is closed. (You can construct open intervals in the usual topology on R by a countably infinite intersection of closed sets - you may want to prove that...).

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