Question:
Consider the metric spaces \( X = (\mathbb{R}, d_1) \) and \( Y = ([0, 1], d_2) \) with the metrics defined by
\[
d_1(x, y) = |x - y|, \quad x, y \in \mathbb{R}, \quad {and} \quad d_2(x, y) = |x - y|, \quad x, y \in [0, 1],
\]
respectively. Then, which one of the following is TRUE?
Consider the metric spaces \( X = (\mathbb{R}, d_1) \) and \( Y = ([0, 1], d_2) \) with the metrics defined by
\[
d_1(x, y) = |x - y|, \quad x, y \in \mathbb{R}, \quad {and} \quad d_2(x, y) = |x - y|, \quad x, y \in [0, 1],
\]
respectively. Then, which one of the following is TRUE?
Show Hint
In metric spaces, for a subspace, a set is open if it can be written as the intersection of an open set in the parent space with the subspace. For \( [0, \frac{1}{4}] \) in \( Y \), this is true.
Updated On: Apr 9, 2025
- \( [0, \frac{1}{4}] \) is open in \( X \) but not in \( Y \)
- \( [0, \frac{1}{4}] \) is open in \( Y \) but not in \( X \)
- \( [0, \frac{1}{4}] \) is open in both \( X \) and \( Y \)
- \( [0, \frac{1}{4}] \) is open neither in \( X \) nor in \( Y \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Spaces \( X \) and \( Y \)
We are given two metric spaces:
\( X = (\mathbb{R}, d_1) \) with the Euclidean metric \( d_1(x, y) = |x - y| \).
\( Y = ([0, 1], d_2) \) with the Euclidean metric restricted to the interval \( [0, 1] \), where \( d_2(x, y) = |x - y| \) for \( x, y \in [0, 1] \).
Step 2: Openness in \( X \)
In the metric space \( X = (\mathbb{R}, d_1) \), the set \( [0, \frac{1}{4}] \) is not open because it includes its boundary points (0 and \( \frac{1}{4} \)), and in \( \mathbb{R} \), intervals containing their endpoints are not open. So, \( [0, \frac{1}{4}] \) is not open in \( X \).
Step 3: Openness in \( Y \)
In the metric space \( Y = ([0, 1], d_2) \), the set \( [0, \frac{1}{4}] \) is open in \( Y \). Since \( Y \) is a subspace of \( \mathbb{R} \), an open set in \( Y \) is one that can be expressed as an intersection of an open set in \( \mathbb{R} \) with \( Y \). The interval \( (0, \frac{1}{2}) \) is open in \( \mathbb{R} \), and its intersection with \( Y \) can give sets like \( [0, \frac{1}{4}] \) considered open in the subspace topology under suitable base open sets.
Step 4: Conclusion
Therefore, the correct answer is \( \boxed{B} \), since \( [0, \frac{1}{4}] \) is open in \( Y \) but not in \( X \).
Final Answer
\[ \boxed{B} \quad [0, \frac{1}{4}] \text{ is open in } Y \text{ but not in } X \]
We are given two metric spaces:
\( X = (\mathbb{R}, d_1) \) with the Euclidean metric \( d_1(x, y) = |x - y| \).
\( Y = ([0, 1], d_2) \) with the Euclidean metric restricted to the interval \( [0, 1] \), where \( d_2(x, y) = |x - y| \) for \( x, y \in [0, 1] \).
Step 2: Openness in \( X \)
In the metric space \( X = (\mathbb{R}, d_1) \), the set \( [0, \frac{1}{4}] \) is not open because it includes its boundary points (0 and \( \frac{1}{4} \)), and in \( \mathbb{R} \), intervals containing their endpoints are not open. So, \( [0, \frac{1}{4}] \) is not open in \( X \).
Step 3: Openness in \( Y \)
In the metric space \( Y = ([0, 1], d_2) \), the set \( [0, \frac{1}{4}] \) is open in \( Y \). Since \( Y \) is a subspace of \( \mathbb{R} \), an open set in \( Y \) is one that can be expressed as an intersection of an open set in \( \mathbb{R} \) with \( Y \). The interval \( (0, \frac{1}{2}) \) is open in \( \mathbb{R} \), and its intersection with \( Y \) can give sets like \( [0, \frac{1}{4}] \) considered open in the subspace topology under suitable base open sets.
Step 4: Conclusion
Therefore, the correct answer is \( \boxed{B} \), since \( [0, \frac{1}{4}] \) is open in \( Y \) but not in \( X \).
Final Answer
\[ \boxed{B} \quad [0, \frac{1}{4}] \text{ is open in } Y \text{ but not in } X \]
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