Proposition 4.4.6. A subset A of a metric space S is closed if and only if A' C A. When we considered the set A = (0,1) in R, we observed that A' = [0, 1). Clearly A' is not contained in A, thus by Proposition 4.4.6, A = (0, 1) is not closed. For a, b €R with a

Elementary Linear Algebra (MindTap Course List)
8th Edition
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Author:Ron Larson
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Chapter4: Vector Spaces
Section4.2: Vector Spaces
Problem 38E: Determine whether the set R2 with the operations (x1,y1)+(x2,y2)=(x1x2,y1y2) and c(x1,y1)=(cx1,cy1)...
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**Proposition 4.4.6.** A subset \( A \) of a metric space \( S \) is closed if and only if \( A' \subseteq A \).

When we considered the set \( A = (0, 1) \) in \( \mathbb{R} \), we observed that \( A' = [0, 1] \). Clearly \( A' \) is not contained in \( A \), thus by Proposition 4.4.6, \( A = (0, 1) \) is not closed.

*For \( a, b \in \mathbb{R} \) with \( a < b \), is \( (a, b] \) closed? Explain by appealing to Proposition 4.4.6.*
Transcribed Image Text:**Proposition 4.4.6.** A subset \( A \) of a metric space \( S \) is closed if and only if \( A' \subseteq A \). When we considered the set \( A = (0, 1) \) in \( \mathbb{R} \), we observed that \( A' = [0, 1] \). Clearly \( A' \) is not contained in \( A \), thus by Proposition 4.4.6, \( A = (0, 1) \) is not closed. *For \( a, b \in \mathbb{R} \) with \( a < b \), is \( (a, b] \) closed? Explain by appealing to Proposition 4.4.6.*
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