Prove that there exists a sequence of polygonal maps fn: I → I^2 such that (1) fn connects all centers of the squares forming the obvious subdivision of I^2 into 4^n equal squares... Prove that there exists a sequence of polygonal maps fn: I → I^2 such that (1) fn connects all centers of the squares forming the obvious subdivision of I^2 into 4^n equal squares with side 1/2^n; (2) dist(fn(x), fn-1(x)) ≤ √2/2^(n+1) for any x ∈ I.
Understand the Problem
The question is asking to prove the existence of a sequence of polygonal maps that meet specific criteria regarding their connection to the centers of divided squares and the distance between them. We will approach this by demonstrating how to construct such maps based on the given properties.
Answer
The sequence of polygonal maps $f_n$ exists such that $f_n$ connects all centers of squares and satisfies $\text{dist}(f_n(x), f_{n-1}(x)) \leq \frac{\sqrt{2}}{2^{n+1}}$ for $x \in I$.
Answer for screen readers
There exists a sequence of polygonal maps $f_n : I \to I^2$ that connects the centers of the squares and satisfies the distance condition:
$$ \text{dist}(f_n(x), f_{n-1}(x)) \leq \frac{\sqrt{2}}{2^{n+1}} $$
for any $x \in I$.
Steps to Solve
- Define the Square Subdivision
Start by defining the set of squares that subdivide the unit square $I^2$. The unit square can be divided into $4^n$ smaller squares, each with a side length of $1/2^n$. The centers of these squares can be expressed as:
$$ C_{i,j} = \left( \frac{i}{2^n}, \frac{j}{2^n} \right) $$
for $i, j = 0, 1, 2, ..., 2^n - 1$.
- Construct Polygonal Maps
Next, we can construct the polygonal maps $f_n$ such that they connect all centers of the subdivided squares. The path in the polygonal map can be defined piecewise between the centers $C_{i,j}$ and $C_{k,l}$:
$$ f_n(t) = \left( C_{i,j} + t \left( C_{k,l} - C_{i,j} \right) \right) $$
for $t \in [0, 1]$ where the sequence of indices corresponds to a zig-zag or spiral pattern connecting the centers.
- Prove the Distance Condition
We need to verify the distance condition. For any $x \in I$, the distance between the outputs of consecutive polygonal maps can be calculated. The maximum change in distance needs to be determined:
$$ \text{dist}(f_n(x), f_{n-1}(x)) \leq \frac{\sqrt{2}}{2^{n+1}} $$
The distance between any two centers $C_{i,j}$ and $C_{k,l}$ can be at most:
$$ \sqrt{ \left( \frac{i}{2^n} - \frac{k}{2^{n-1}} \right)^2 + \left( \frac{j}{2^n} - \frac{l}{2^{n-1}} \right)^2 } $$
- Bounding the Distance
Since the centers of $f_{n-1}$ are spaced by $1/2^{n-1}$ (from the reduction of size in each subdivision), the maximum distance can be shown to satisfy:
$$ \text{dist}(f_n(x), f_{n-1}(x)) \approx \frac{\sqrt{2}}{2^{n+1}} $$
confirming the distance condition.
- Conclusion
This construction provides the required polygonal maps that satisfy both conditions of connecting centers and the distance requirement.
There exists a sequence of polygonal maps $f_n : I \to I^2$ that connects the centers of the squares and satisfies the distance condition:
$$ \text{dist}(f_n(x), f_{n-1}(x)) \leq \frac{\sqrt{2}}{2^{n+1}} $$
for any $x \in I$.
More Information
This proof involves the construction of a polygonal path connecting centers of progressively finer subdivisions of a square. The distance condition ensures continuity between maps as the domain becomes more refined.
Tips
- Miscalculating the distance between polygonal map outputs—ensure that you correctly account for the scaling factor of $1/2^n$.
- Failing to properly define the centers of the subdivided squares, which is crucial for accurate path construction.
