Let $X$ and $Y$ be compact metric spaces, $T:X \to X$ and $S: Y \to Y$ homeomorphisms. We say $T$ and $S$ are topologically conjugate if there exists a homeomorphism $\phi: X \to Y$ such that $S(\phi(x))= \phi(T(x))$ for all $x \in X$.

We say a homeomorphism $T$ is minimal if for every $x \in X$, the set $\{T^k x: x\in \mathbb{Z}\}$ is dense in $X$.

The claim is that T is minimal iff S is minimal.

I have tried to show this but to no avail, as I get stuck trying to compare a ball around x with a ball around $\phi(x)$ which unless its an isometry I can not do anything interesting with to get my result. Thanks in advance!

  • $\begingroup$ Your terminology is inaccurate, it should be: "We say $T$ and $S$ [not $X$ and $Y$] are topologically conjugate if there exists $\phi : X \to Y$..." $\endgroup$ – Lee Mosher Aug 8 '17 at 23:45
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    $\begingroup$ Also, please use MathJax for the mathematical formulas math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Lee Mosher Aug 8 '17 at 23:46
  • $\begingroup$ Thanks Lee Mosher I have edited the title. I will also check MathJax for future use. $\endgroup$ – user172377 Aug 8 '17 at 23:48
  • $\begingroup$ You don't need to use that $\phi$ is an isometry, you only need the $\epsilon,\delta$ definition of continuity. $\endgroup$ – Lee Mosher Aug 8 '17 at 23:54
  • $\begingroup$ @LeeMosher: Here is why the OP needs an isometry, based on the approach. $\endgroup$ – Cameron Buie Aug 9 '17 at 0:02

It suffices to assume that $T$ is minimal and prove that $S$ is minimal, since the converse follows by the same argument, swapping $(T,S,\phi) \to (S, T, \phi^{-1})$. In metric spaces, we can use sequences.

Take $y \in Y$. We want to prove that $\{ S^ky \mid y \in Y \}$ is dense in $Y$. Let $y_0 \in Y$. Then $\phi^{-1}(y_0) \in X$, and since $T$ is minimal, there is a sequence $(k_n)_{n \geq 0}$ such that $T^{k_n}(\phi^{-1}(y)) \to \phi^{-1}(y_0)$. By continuity of $\phi$, and noting that $\phi \circ T \circ \phi^{-1} = S$ implies $\phi \circ T^{k_n} \circ \phi^{-1} = S^{k_n}$, we obtain $S^{k_n}(y) \to y_0$. Done.

  • $\begingroup$ Thanks for the great explanation! $\endgroup$ – user172377 Aug 9 '17 at 3:59

It suffices to prove that if $T$ is minimal, then $S$ is minimal. Suppose $\{T^k(x) \mid k\in\mathbb{Z}\}$ is a dense subset of $X$ for each $x\in X$. Fix $y\in Y$. To show that $\{S^k(y) \mid k\in\mathbb{Z}\}$ is dense in $Y$, we let $U$ be a nonempty subset of $Y$. Then $\phi^{-1}(U)$ is a nonempty open subset of $X$, so by the density of $\{T^k(\phi^{-1}(y)) \mid k\in\mathbb{Z}\}$, there exists $k\in\mathbb{Z}$ such that $T^k(\phi^{-1}(y))\in\phi^{-1}(U)$. Therefore $$S^k(y)=S^k(\phi(\phi^{-1}(y)))=\phi(T^k(\phi^{-1}(y)))\in U,$$ as desired.

This shows that not only do we not need $\phi$ to be an isometry, but we don't even require $X$ and $Y$ to be compact or metrizable.

  • $\begingroup$ Thanks! Just to make sure, proving it for any open set is enough since we simply need to show we can approximate any point by an iterate of some other point, which is the same as showing any open ball around that iterated point will contain the other point? $\endgroup$ – user172377 Aug 9 '17 at 4:00
  • $\begingroup$ That's essentially the idea. In other words, a point $x$ is in the closure of $A$ iff every open set containing $x$ intersects $A$. This gives the characterization that a set $D$ is dense in $X$ iff every nonempty open set in $X$ intersects $D$. $\endgroup$ – John Griffin Aug 9 '17 at 4:03
  • $\begingroup$ (+1) for pointing out that compactness and metrizability are irrelevant here. $\endgroup$ – Adayah Aug 9 '17 at 8:56

The fact you need, which is true for homeomorphisms even when they are not isometries, is this:

  • for any homeomorphism $\phi : X \to Y$, a subset $A \subset X$ is dense if and only if the image subset $\phi(A) \subset Y$ is dense.

As the other answers show, if $y = \phi(x)$ you can apply this fact using $A = \{T^k(x) \,|\, k \in \mathbb{Z}\}$ and $$\phi(A) = \{\phi(T^k(x)) \,|\, k \in \mathbb{Z}\} = \{S^k(\phi(x)) \, | \, k \in \mathbb{Z}\} = \{T^k(y) \,|\, k \in \mathbb{Z}\} $$

You don't need that the homeomorphism is an isometry to prove this fact. You can use the open sets definition of continuity, as the other answers show.

Or, you can use the $\epsilon,\delta$ definition, in the following manner.

Suppose we know that $A \subset X$ is dense. Let's prove that $\phi(A) \subset Y$ is dense. Consider $y \in Y$ and $\epsilon>0$ and the open ball $B(y,\epsilon) \subset Y$. We must find a point of $\phi(A)$ contained in the ball $B(y,\epsilon)$. Let $x = \phi^{-1}(y)$. By continuity of $\phi$, there exists $\delta>0$ such that $\phi(B(x,\delta)) \subset B(y,\epsilon)$. Since $A \subset X$ is dense, there exists $a \in A$ such that $a \in B(x,\delta)$. It follows that $\phi(a) \in \phi(B(x,\delta)) \subset B(y,\epsilon)$, and clearly $\phi(a) \in \phi(A)$.

  • $\begingroup$ I think you meant $\{S^k(\phi(x))\mid k\in\Bbb Z\}$, etc, when computing $\phi(A)$ in the first part of the answer, no? $\endgroup$ – Ivo Terek Aug 9 '17 at 2:47
  • $\begingroup$ Thanks for the correction @IvoTerek $\endgroup$ – Lee Mosher Aug 9 '17 at 13:27

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