证明存在拓扑共轭：两类二维子集映射的拓扑共轭性求证

Let's break this down step by step—we need to show there's a homeomorphism between spaces $Y$ and $X$ that makes the maps $f$ and $g$ commute (the core definition of topological conjugacy). First, let's recap the setup clearly to avoid confusion:

Key Definitions #

• Space $X$: A subset of $\mathbb{R}^2$ given by $X = \bigcup_{n \in \mathbb{Z}} X_n$, where each $X_n = {n} \times [0, 2^{-|n|}]$. It inherits the standard Euclidean metric from $\mathbb{R}^2$.
• Map $f: X \to X$: Defined piecewise:
  $$f(n,y) = \begin{cases}
  (n+1, 2y) & \text{if } n < 0, \
  (n+1, \frac{y}{2}) & \text{if } n \geq 0.
  \end{cases}$$
• Space $Y$: Another subset of $\mathbb{R}^2$: $Y = \bigcup_{n \in \mathbb{Z}} {n} \times [0,1]$, also with the Euclidean subspace metric.
• Map $g: Y \to Y$: A simple shift on the first component: $g(n,y) = (n+1, y)$ (the second component stays unchanged).

Topological conjugacy requires a homeomorphism $h: Y \to X$ such that $f \circ h = h \circ g$. Let's construct this $h$ and verify all necessary conditions.

Step 1: Construct the Homeomorphism $h$ #

Define $h: Y \to X$ by scaling the second component of each point in $Y$ to fit the corresponding $X_n$:
$$h(n,y) = \left(n, y \cdot 2^{-|n|}\right)$$
for every $(n,y) \in Y$ (so $n \in \mathbb{Z}$, $y \in [0,1]$).

Verify $h$ is a Homeomorphism #

We need to confirm $h$ is bijective, continuous, and has a continuous inverse:

1. Injective: Suppose $h(n_1,y_1) = h(n_2,y_2)$. Then $n_1 = n_2$ (since first components match), so $y_1 \cdot 2^{-|n_1|} = y_2 \cdot 2^{-|n_1|}$. Since $2^{-|n_1|} > 0$, we can divide both sides to get $y_1 = y_2$. Thus $h$ is injective.
2. Surjective: Take any $(n,y) \in X$. By definition of $X$, $y \in [0, 2^{-|n|}]$. Let $y' = y \cdot 2^{|n|}$—this gives $y' \in [0,1]$, so $(n,y') \in Y$, and $h(n,y') = (n, y' \cdot 2^{-|n|}) = (n,y)$. Thus $h$ is surjective.
3. Continuous: Since $X$ and $Y$ are subspaces of $\mathbb{R}^2$, we use sequential continuity. Suppose $(n_k, y_k) \to (n,y)$ in $Y$. The integer component $n_k$ must eventually equal $n$ (since integer parts are discrete in $\mathbb{R}^2$), and $y_k \to y$. Then $h(n_k,y_k) = (n_k, y_k \cdot 2^{-|n_k|}) \to (n, y \cdot 2^{-|n|}) = h(n,y)$, so $h$ is continuous.
4. Continuous Inverse: The inverse map $h^{-1}: X \to Y$ is given by $h^{-1}(n,y) = (n, y \cdot 2^{|n|})$. Using sequential continuity again: if $(n_k,y_k) \to (n,y)$ in $X$, then $n_k$ eventually equals $n$, $y_k \to y$, so $h^{-1}(n_k,y_k) = (n_k, y_k \cdot 2^{|n_k|}) \to (n, y \cdot 2^{|n|}) = h^{-1}(n,y)$. Thus $h^{-1}$ is continuous.

All conditions are satisfied, so $h$ is a homeomorphism between $Y$ and $X$.

Step 2: Verify the Conjugacy Condition $f \circ h = h \circ g$ #

We check this for both cases of $n$:

Case 1: $n \geq 0$ #

• Left-hand side (LHS): $f(h(n,y)) = f\left(n, y \cdot 2^{-n}\right)$. Since $n \geq 0$, $f$ maps $(n,z)$ to $(n+1, z/2)$. Substituting $z = y \cdot 2^{-n}$, we get:
  $$f(h(n,y)) = \left(n+1, \frac{y \cdot 2^{-n}}{2}\right) = \left(n+1, y \cdot 2^{-(n+1)}\right)$$
• Right-hand side (RHS): $h(g(n,y)) = h(n+1,y) = \left(n+1, y \cdot 2^{-|n+1|}\right)$. Since $n+1 \geq 1$, $|n+1| = n+1$, so this simplifies to $\left(n+1, y \cdot 2^{-(n+1)}\right)$.
• LHS = RHS, so the condition holds here.

Case 2: $n < 0$ #

• Left-hand side (LHS): $f(h(n,y)) = f\left(n, y \cdot 2^{-|n|}\right)$. Since $n < 0$, $|n| = -n$, so $2^{-|n|} = 2^{n}$. $f$ maps $(n,z)$ to $(n+1, 2z)$, so substituting $z = y \cdot 2^{n}$:
  $$f(h(n,y)) = \left(n+1, 2 \cdot y \cdot 2^{n}\right) = \left(n+1, y \cdot 2^{n+1}\right)$$
• Right-hand side (RHS): $h(g(n,y)) = h(n+1,y)$. If $n = -1$, then $n+1 = 0$, so $h(0,y) = (0, y \cdot 2^{0}) = (0,y)$, which matches LHS (since $f(h(-1,y)) = f(-1, y \cdot 2^{-1}) = (0, 2 \cdot y \cdot 2^{-1}) = (0,y)$). For $n < -1$, $n+1 < 0$, so $|n+1| = -(n+1)$, and $2^{-|n+1|} = 2^{n+1}$. Thus $h(n+1,y) = \left(n+1, y \cdot 2^{n+1}\right)$, which equals LHS.
• LHS = RHS, so the condition holds here too.

Conclusion #

We've constructed a homeomorphism $h: Y \to X$ that satisfies $f \circ h = h \circ g$, so $f$ and $g$ are topologically conjugate.

内容的提问来源于stack exchange，提问作者user479859