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June 20th, 2016, 01:38 PM   #1
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Connectedness - Relatively Close

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As an example of a subset of the real line that is disconnected, let A = [0,1] ∪ (2,3). [0, 1] is a relatively closed subset of A since [0, 1] is closed in R. At the same time [0, 1] is a relatively open subset of A, since [0, 1] = (-1/2, 3/2) ∩ A . Finally, [0, 1] ≠ Ø and [0, 1] ≠ A, hence A is disconnected. By the same token, the “open interval” (2, 3) is also both relatively open and relatively closed in A.
I don't understand why is it that since [0, 1] = (-1/2, 3/2) ∩ A, [0, 1] is a relatively open subset of A.

Isn't it that the intersection of (-1/2, 3/2) and A is again [0, 1] is a closed subset of A?

Wonder if someone could help point that out to me.

Thank you!!
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June 22nd, 2016, 05:49 AM   #2
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Given A a subset of X, a set U is "open in A" (or "relative to A") if and only if U is equal to the intersection of an open subset of X with A.

Similarly, U is "closed in A" if and only if U is equal to the intersection of a closed subset of X with A.

(Sometimes those are used as the definitions of "open in A" and "closed in A".)
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