Question:
The set
\[
\left\{ \frac{x}{1+x^2} : -1<x<1 \right\}, \text{ as a subset of } \mathbb{R}, \text{ is}
\]
The set
\[
\left\{ \frac{x}{1+x^2} : -1<x<1 \right\}, \text{ as a subset of } \mathbb{R}, \text{ is}
\]
Show Hint
A set is compact if it is closed and bounded. In this case, the set is bounded but not closed.
Updated On: Dec 11, 2025
- connected and compact
- connected but not compact
- not connected but compact
- neither connected nor compact
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the set.
The set \( \left\{ \frac{x}{1+x^2} : -1<x<1 \right\} \) is defined by a continuous function \( f(x) = \frac{x}{1 + x^2} \), where \( x \) lies between \( -1 \) and \( 1 \). This is a continuous function on the open interval \( (-1, 1) \), and as a result, the image of this function will be a connected set.
Step 2: Analyzing compactness.
The set is not compact because it is not closed. The function \( f(x) \) does not include the boundary points \( -1 \) and \( 1 \) in its image, making the set non-compact.
Step 3: Conclusion.
Thus, the set is connected but not compact, so the correct answer is \( \boxed{(B)} \).
The set \( \left\{ \frac{x}{1+x^2} : -1<x<1 \right\} \) is defined by a continuous function \( f(x) = \frac{x}{1 + x^2} \), where \( x \) lies between \( -1 \) and \( 1 \). This is a continuous function on the open interval \( (-1, 1) \), and as a result, the image of this function will be a connected set.
Step 2: Analyzing compactness.
The set is not compact because it is not closed. The function \( f(x) \) does not include the boundary points \( -1 \) and \( 1 \) in its image, making the set non-compact.
Step 3: Conclusion.
Thus, the set is connected but not compact, so the correct answer is \( \boxed{(B)} \).
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