It is known that if continuous action $\varphi:G\times S^1\to S^1$ is minimal, then $\varphi$ is either equicontinuous or for every $x\in S^1$ there exist open interval $I$ of $x$ and $\{g_n\}\subseteq G$ such that the length of the intervals $g_n(I)$ convergence to zero.
Note that $\varphi$ is minimal if orbit of every point of $S^1$ is dense. Also $\varphi$ is equicontinous if for every $\epsilon>0$ there is $\delta>0$ such that if $d(x, y)<\delta$ then for all $g\in G$ we have $d(\varphi(g, x), \varphi(g, y))<\epsilon$.
In my research I want to prove the following:
Let $\varphi:G\times S^1\to S^1$ be minimal action. Then the following statement are equivalent:
1)for every $x\in S^1$ there exist open interval $I$ of $x$ and $\{g_n\}\subseteq G$ such that the length of the intervals $g_n(I)$ convergence to zero,
2) there is $\delta>0$ such that if $x\neq y$ then there is $g\in G$ with $d(\varphi(g, x), \varphi(g, y))>\delta$
It is clear that item 2 implies that $\varphi$ is not equicontinuous, hence if $\varphi$ is minimal, then item (1) is hold. Thus item (2) implies item (1). But I can not prove $1\to 2$. Would you please to help me to know it?
Thanks a lot