Question:
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:
P: \(X\) is a connected subset of \(\mathbb{R}^2\).
Q: \(Y\) is a connected subset of \(\mathbb{R}^2\).
Then
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:
P: \(X\) is a connected subset of \(\mathbb{R}^2\).
Q: \(Y\) is a connected subset of \(\mathbb{R}^2\).
Then
\[ X = \left\{\left(x, x \sin\frac{1}{x}\right) \in \mathbb{R}^2 : x \in (0,1]\right\} \cup \{(0, y) \in \mathbb{R}^2 : -\infty < y < \infty\}, \]
\[ Y = \left\{\left(x, \sin\frac{1}{x}\right) \in \mathbb{R}^2 : x \in (0,1]\right\} \cup \{(0, y) \in \mathbb{R}^2 : -\infty < y < \infty\}. \]
Consider the following statements:
P: \(X\) is a connected subset of \(\mathbb{R}^2\).
Q: \(Y\) is a connected subset of \(\mathbb{R}^2\).
Then
Show Hint
To prove a set \(S = S_1 \cup S_2\) is connected, a powerful strategy is to:
1. Show \(S_1\) is connected.
2. Consider its closure, \(\bar{S_1}\), which is also connected.
3. Show that \(S_2\) is connected.
4. If \(\bar{S_1}\) and \(S_2\) have a point in common, their union is connected. This often works when one part of the set contains the limit points of another part.
Updated On: Sep 5, 2025
- both P and Q are TRUE
- P is FALSE and Q is TRUE
- P is TRUE and Q is FALSE
(D) both P and Q are FALSE
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This question tests the concept of connectedness in topology. A key theorem states that if A is a connected set, then any set C such that \(A \subseteq C \subseteq \bar{A}\) (the closure of A) is also connected. We will analyze the connectedness of sets X and Y by identifying their component parts and examining their closures.
Step 3: Detailed Explanation:
Analysis of Statement P:
Let \(A = \left\{\left(x, x \sin\frac{1}{x}\right) : x \in (0,1]\right\}\) and \(B = \{(0, y) : -\infty<y<\infty\}\) (the y-axis). So \(X = A \cup B\).
The set A is the graph of a continuous function \(g(x) = x\sin(1/x)\) on the connected interval \((0,1]\). The image of a connected set under a continuous map is connected, so A is connected.
The set B (the y-axis) is also clearly connected.
Now let's find the closure of A, \(\bar{A}\). As \(x \to 0^+\), we have \(|\sin(1/x)| \le 1\), so \(|x\sin(1/x)| \le |x|\). By the Squeeze Theorem, \(\lim_{x \to 0^+} x\sin(1/x) = 0\). This means the graph of A approaches the origin (0,0) as \(x\) approaches 0. The set of limit points of A is \(A\) itself plus the point (0,0). So, \(\bar{A} = A \cup \{(0,0)\}\).
Since A is connected, its closure \(\bar{A}\) is also connected. The set X is the union of two connected sets: \(\bar{A}\) and B. These two sets are not disjoint; they share the point (0,0). The union of two connected sets with a non-empty intersection is connected.
Therefore, \(X = \bar{A} \cup B\) is a connected set.
Thus, P is TRUE.
Analysis of Statement Q:
Let \(A' = \left\{\left(x, \sin\frac{1}{x}\right) : x \in (0,1]\right\}\) and \(B = \{(0, y) : -\infty<y<\infty\}\). So \(Y = A' \cup B\).
The set A' (the topologist's sine curve graph) is the graph of a continuous function on \((0,1]\), so it is connected.
Let's find the closure of A', \(\bar{A'}\). As \(x \to 0^+\), the term \(\sin(1/x)\) oscillates infinitely often between -1 and 1. This means that for any \(y_0 \in [-1, 1]\), the point \((0, y_0)\) is a limit point of A'. Let \(L = \{(0, y) : -1 \le y \le 1\}\). The closure of A' is \(\bar{A'} = A' \cup L\).
Since A' is connected, its closure \(\bar{A'}\) is also connected.
The set Y is the union of two connected sets: \(\bar{A'}\) and B (the y-axis). Their intersection is \(\bar{A'} \cap B = L\). Since the intersection is non-empty, their union \(Y = \bar{A'} \cup B\) is connected.
Thus, Q is TRUE.
Step 4: Final Answer:
Both statements P and Q are TRUE.
Step 5: Why This is Correct:
In both cases, the set can be seen as the union of two connected subsets that have a non-empty intersection. For X, the graph part connects to the y-axis at the origin. For Y, the graph part's closure includes a segment of the y-axis, ensuring a connection. The union of connected sets with a non-empty intersection is always connected.
This question tests the concept of connectedness in topology. A key theorem states that if A is a connected set, then any set C such that \(A \subseteq C \subseteq \bar{A}\) (the closure of A) is also connected. We will analyze the connectedness of sets X and Y by identifying their component parts and examining their closures.
Step 3: Detailed Explanation:
Analysis of Statement P:
Let \(A = \left\{\left(x, x \sin\frac{1}{x}\right) : x \in (0,1]\right\}\) and \(B = \{(0, y) : -\infty<y<\infty\}\) (the y-axis). So \(X = A \cup B\).
The set A is the graph of a continuous function \(g(x) = x\sin(1/x)\) on the connected interval \((0,1]\). The image of a connected set under a continuous map is connected, so A is connected.
The set B (the y-axis) is also clearly connected.
Now let's find the closure of A, \(\bar{A}\). As \(x \to 0^+\), we have \(|\sin(1/x)| \le 1\), so \(|x\sin(1/x)| \le |x|\). By the Squeeze Theorem, \(\lim_{x \to 0^+} x\sin(1/x) = 0\). This means the graph of A approaches the origin (0,0) as \(x\) approaches 0. The set of limit points of A is \(A\) itself plus the point (0,0). So, \(\bar{A} = A \cup \{(0,0)\}\).
Since A is connected, its closure \(\bar{A}\) is also connected. The set X is the union of two connected sets: \(\bar{A}\) and B. These two sets are not disjoint; they share the point (0,0). The union of two connected sets with a non-empty intersection is connected.
Therefore, \(X = \bar{A} \cup B\) is a connected set.
Thus, P is TRUE.
Analysis of Statement Q:
Let \(A' = \left\{\left(x, \sin\frac{1}{x}\right) : x \in (0,1]\right\}\) and \(B = \{(0, y) : -\infty<y<\infty\}\). So \(Y = A' \cup B\).
The set A' (the topologist's sine curve graph) is the graph of a continuous function on \((0,1]\), so it is connected.
Let's find the closure of A', \(\bar{A'}\). As \(x \to 0^+\), the term \(\sin(1/x)\) oscillates infinitely often between -1 and 1. This means that for any \(y_0 \in [-1, 1]\), the point \((0, y_0)\) is a limit point of A'. Let \(L = \{(0, y) : -1 \le y \le 1\}\). The closure of A' is \(\bar{A'} = A' \cup L\).
Since A' is connected, its closure \(\bar{A'}\) is also connected.
The set Y is the union of two connected sets: \(\bar{A'}\) and B (the y-axis). Their intersection is \(\bar{A'} \cap B = L\). Since the intersection is non-empty, their union \(Y = \bar{A'} \cup B\) is connected.
Thus, Q is TRUE.
Step 4: Final Answer:
Both statements P and Q are TRUE.
Step 5: Why This is Correct:
In both cases, the set can be seen as the union of two connected subsets that have a non-empty intersection. For X, the graph part connects to the y-axis at the origin. For Y, the graph part's closure includes a segment of the y-axis, ensuring a connection. The union of connected sets with a non-empty intersection is always connected.
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