Final answer:
The Cantor set Γ is homeomorphically mapped onto itself by the linear map L(x)=3x when applied to Γ ∩ [0, 1/3], due to the self-similar recursive nature of Γ and the continuity and bijectiveness of the map with continuous inverse.
Step-by-step explanation:
The question asks to prove that the linear map L(x)=3x maps the intersection of the Cantor set Γ and the interval [0, 1/3] homeomorphically onto Γ. The Cantor set is a classic example of a fractal, constructed by repeatedly removing the middle third of a line segment. To approach this proof, we can observe that:
- The Cantor set is defined by iteratively removing the open middle third from each closed interval in the previous step.
- Γ ∩ [0, 1/3] consists of all points in the Cantor set that are in the interval [0, 1/3].
- The map L(x) expands each point in [0, 1/3] by a factor of 3.
- Each step in the construction of Γ is self-similar, and the portion of Γ in [0, 1/3] is a scaled-down version of the entire set Γ.
Mapping Γ ∩ [0, 1/3] by L(x) effectively stretches this portion to cover the entire original interval [0, 1] from which Γ is constructed. Since L is a continuous bijection whose inverse is also continuous (scaling by 1/3), it is a homeomorphism. Thus, each point in Γ ∩ [0, 1/3] corresponds uniquely to a point in Γ, and the topological structure is preserved. This demonstrates that L(x)=3x maps Γ ∩ [0, 1/3] homeomorphically onto Γ.