If T and S are topologically conjugate homeomorphisms, T on X is minimal iff S on Y is minimal. (Definitions inside) 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!
 A: 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.
A: 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.
A: 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 $\phi(a) \in \phi(A)$.
