Question:
Let \(\mathbb{R}^3\) be a topological space with the usual topology and \(\mathbb{Q}\) denote the set of rational numbers. Define the subspaces X, Y, Z and W of \(\mathbb{R}^3\) as follows:
\(X = \{(x, y, z) \in \mathbb{R}^3 : |x| + |y| + |z| \in \mathbb{Q}\}\)
\(Y = \{(x, y, z) \in \mathbb{R}^3 : xyz = 1\}\)
\(Z = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}\)
\(W = \{(x, y, z) \in \mathbb{R}^3 : xyz = 0\}\)
Which of the following statements is correct?
Let \(\mathbb{R}^3\) be a topological space with the usual topology and \(\mathbb{Q}\) denote the set of rational numbers. Define the subspaces X, Y, Z and W of \(\mathbb{R}^3\) as follows:
\(X = \{(x, y, z) \in \mathbb{R}^3 : |x| + |y| + |z| \in \mathbb{Q}\}\)
\(Y = \{(x, y, z) \in \mathbb{R}^3 : xyz = 1\}\)
\(Z = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}\)
\(W = \{(x, y, z) \in \mathbb{R}^3 : xyz = 0\}\)
Which of the following statements is correct?
\(X = \{(x, y, z) \in \mathbb{R}^3 : |x| + |y| + |z| \in \mathbb{Q}\}\)
\(Y = \{(x, y, z) \in \mathbb{R}^3 : xyz = 1\}\)
\(Z = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}\)
\(W = \{(x, y, z) \in \mathbb{R}^3 : xyz = 0\}\)
Which of the following statements is correct?
Show Hint
To quickly disprove a homeomorphism, check for fundamental topological invariants:
1. Connectedness (and number of connected components).
2. Compactness.
3. Simple connectedness (presence of "holes").
If any of these differ, the spaces cannot be homeomorphic.
Updated On: Sep 5, 2025
- X is homeomorphic to Y
- Z is homeomorphic to W
- Y is homeomorphic to W
- X is NOT homeomorphic to W
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Two topological spaces are homeomorphic if there exists a continuous bijection between them with a continuous inverse. Homeomorphic spaces share all topological properties, such as connectedness, compactness, path-connectedness, etc. We can prove two spaces are NOT homeomorphic by finding a topological property that one space has but the other does not.
Step 3: Detailed Explanation:
Let's analyze the topological properties of each space.
- X: This space consists of points whose taxicab norm is a rational number. Between any two distinct points in X, we can find a point whose taxicab norm is irrational. This means there is no path connecting any two points within X. Therefore, X is a totally disconnected space.
- Y: This surface consists of four separate, disconnected components (one in each octant where the product of coordinates is positive: (+,+,+), (+,-,-), (-,+,-), (-,,-,+)). Each component is a smooth, path-connected surface. So, Y is not connected.
- Z: This is the unit sphere \(S^2\). It is compact, connected, and path-connected.
- W: This is the union of the three coordinate planes (\(x=0\), \(y=0\), \(z=0\)). Any point on one plane can be connected to any point on another plane via a path that passes through the origin. Therefore, W is connected and path-connected.
Now let's evaluate the options:
(A) X is homeomorphic to Y: X is totally disconnected. Y is not (it consists of four connected components which are surfaces). Thus, they are not homeomorphic.
(B) Z is homeomorphic to W: Z (the sphere) is compact. W (the union of three infinite planes) is not compact. Since compactness is a topological invariant, they are not homeomorphic.
(C) Y is homeomorphic to W: Y has four connected components. W is a single connected component. The number of connected components is a topological invariant. Thus, they are not homeomorphic.
(D) X is NOT homeomorphic to W: X is totally disconnected. W is connected. Since connectedness is a topological invariant and one space possesses it while the other does not, they cannot be homeomorphic. Therefore, the statement that they are NOT homeomorphic is TRUE.
Step 4: Final Answer:
The correct statement is that X is NOT homeomorphic to W.
Step 5: Why This is Correct:
The proof relies on identifying a fundamental topological property—connectedness—that differs between the spaces X and W. X is totally disconnected, while W is connected. This difference makes a homeomorphism between them impossible.
Two topological spaces are homeomorphic if there exists a continuous bijection between them with a continuous inverse. Homeomorphic spaces share all topological properties, such as connectedness, compactness, path-connectedness, etc. We can prove two spaces are NOT homeomorphic by finding a topological property that one space has but the other does not.
Step 3: Detailed Explanation:
Let's analyze the topological properties of each space.
- X: This space consists of points whose taxicab norm is a rational number. Between any two distinct points in X, we can find a point whose taxicab norm is irrational. This means there is no path connecting any two points within X. Therefore, X is a totally disconnected space.
- Y: This surface consists of four separate, disconnected components (one in each octant where the product of coordinates is positive: (+,+,+), (+,-,-), (-,+,-), (-,,-,+)). Each component is a smooth, path-connected surface. So, Y is not connected.
- Z: This is the unit sphere \(S^2\). It is compact, connected, and path-connected.
- W: This is the union of the three coordinate planes (\(x=0\), \(y=0\), \(z=0\)). Any point on one plane can be connected to any point on another plane via a path that passes through the origin. Therefore, W is connected and path-connected.
Now let's evaluate the options:
(A) X is homeomorphic to Y: X is totally disconnected. Y is not (it consists of four connected components which are surfaces). Thus, they are not homeomorphic.
(B) Z is homeomorphic to W: Z (the sphere) is compact. W (the union of three infinite planes) is not compact. Since compactness is a topological invariant, they are not homeomorphic.
(C) Y is homeomorphic to W: Y has four connected components. W is a single connected component. The number of connected components is a topological invariant. Thus, they are not homeomorphic.
(D) X is NOT homeomorphic to W: X is totally disconnected. W is connected. Since connectedness is a topological invariant and one space possesses it while the other does not, they cannot be homeomorphic. Therefore, the statement that they are NOT homeomorphic is TRUE.
Step 4: Final Answer:
The correct statement is that X is NOT homeomorphic to W.
Step 5: Why This is Correct:
The proof relies on identifying a fundamental topological property—connectedness—that differs between the spaces X and W. X is totally disconnected, while W is connected. This difference makes a homeomorphism between them impossible.
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