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It R be the metric space then,

A) \(\left( 0,\frac{1}{2} \right]\) is not Open in [0, 2]
B) \(\left( 0,\frac{1}{2} \right]\) is open in [0, 2]
C) \(\left( 0,\frac{1}{2} \right]\) is closed in [0, 2]
D) None of these

Correct Answer:


B) \(\left( 0,\frac{1}{2} \right]\) is open in [0, 2]

Description for Correct answer:
\(\textbf{Result :}\)

Let M be metric space and \(M_{1}\) a subspace of M.

Let \(A_{1}\subseteq M_{1}.\) Then \(A_{1}\) is open in \(M_{1}\) if and only if there exists an open set A in M such that \(A_{1}=A\cap M_{1}.\)

Let \(M = R\)

\(M_{1}=[0,2]\)

Let \(\Large A_{1}= \left[0,\frac{1}{2}\right) \)

Let \(A= \left(\frac{-1}{2},\frac{1}{2}\right) \) which is open in R.

Then \(A_{1}=\left[ 0,\frac{1}{2} \right)= \left(\frac{-1}{2},\frac{1}{2}\right)\cap [0,2] \)

Therefore \(\Large \left[ 0,\frac{1}{2} \right)\) is open in \([0,2]\)

Part of solved Real Analysis questions and answers : >> Elementary Mathematics >> Real Analysis

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