Let \( \mathbb{R} \) denote the set of all real numbers. Consider the following topological spaces.
\[
X_1 = (\mathbb{R}, T_1), \text{where} \, T_1 \text{ is the upper limit topology having all sets } (a, b) \text{ as basis.}
\]
\[
X_2 = (\mathbb{R}, T_2), \text{where} \, T_2 = \{ U \subset \mathbb{R} : \mathbb{R} \setminus U \text{ is finite} \} \cup \{\emptyset\}.
\]
Then:
Show Hint
- both \( X_1 \) and \( X_2 \) are connected
- \( X_1 \) is connected and \( X_2 \) is NOT connected
- \( X_1 \) is NOT connected and \( X_2 \) is connected
- neither \( X_1 \) nor \( X_2 \) is connected
The Correct Option is C
Solution and Explanation
- \( X_2 \) with the topology where the complement of open sets is finite is connected because every set can be separated into open and closed sets with finite complements, allowing a connected space.
Thus, \( X_1 \) is not connected, but \( X_2 \) is connected. Final Answer: (C) \( X_1 \) is NOT connected and \( X_2 \) is connected.
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For each \( n \in \mathbb{N} \), define \( f_n: [0,1] \to \mathbb{R} \) by \( f_n(x) = x^n \) for all \( x \in [0,1] \).
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Which of the following statements is/are correct?
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Define \(f: X \cup Y \to \mathbb{R}\) by \[ f(x) = \begin{cases} 2, & \text{if } x \in X \\ 3, & \text{if } x \in Y \end{cases} \]
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Q: The function \(f: (X \cup Y, \rho_2) \to (\mathbb{R}, \rho_1)\) is uniformly continuous.
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Consider the metric space \((C[0,1], d_\infty)\), where \[ d_\infty(f, g) = \sup\{ |f(x) - g(x)| : x \in [0, 1] \} \text{ for } f, g \in C[0,1]. \] Let \(f_0(x) = 0\) for all \(x \in [0,1]\) and \[ X = \{f \in (C[0, 1], d_\infty) : d_\infty(f_0, f) \ge \frac{1}{2}\}. \] Let \(f_1, f_2 \in C[0, 1]\) be defined by \(f_1(x) = x\) and \(f_2(x) = 1-x\) for all \(x \in [0,1]\).
Consider the following statements:
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Which of the following statements is correct?
- Consider \(\mathbb{R}^2\) with the usual Euclidean metric. Let
\[ X = \left\{\left(x, x \sin\frac{1}{x}\right) \in \mathbb{R}^2 : x \in (0,1]\right\} \cup \{(0, y) \in \mathbb{R}^2 : -\infty < y < \infty\}, \]
\[ Y = \left\{\left(x, \sin\frac{1}{x}\right) \in \mathbb{R}^2 : x \in (0,1]\right\} \cup \{(0, y) \in \mathbb{R}^2 : -\infty < y < \infty\}. \]
Consider the following statements:
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Q: \(Y\) is a connected subset of \(\mathbb{R}^2\).
Then
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