The cluster point set of $\{\frac{1}{n}\sum_{k=0}^{n-1} \delta_{T^kx}\}$ is connected Question: Let $X$ be a compact metric space and $T:X\to X$ is a continuous map. Arbitrarily fixed $x\in X$, let $\delta_{T^kx}$ denote the Dirac measure mass at $T^kx$. Prove The cluster  set of the sequence $\{\frac{1}{n}\sum_{K=0}^{n-1} \delta_{T^kx} \in C(X)^*:n\in \Bbb{N}_{+}\}$ is connected, for the weak-star topology.

My observation: Denote its cluster set by $E$. By Banach-Alaogu theorem, $E$ is a nonempty compact set, and every measure in $E$ is $T$-invariant. But it seems not necessarily being convex. To prove the connection. Suppose $H: E\to \{0,1\}$ be a continuous map, it enough to prove $H$ is constant. But I don’t know how to make $H$ be specific. Another observation is that every nonempty open set of $w^*$-topology is quit big, by definition which contains an finite codimensional subspace.
I also consider some simple examples. Let $X=\mathbb{T}$, with rotation transform, in this case $E$ contains one element-the Lebesgue measure. If $X=[0,1]$, $Tx=x^2$, $E=\{\delta_0\}$ when $x\neq 1$, and $E=\{\delta_1\}$ otherwise. Unfortunately such exampleS don’t give me some useful information because $\#E=1$.
 A: The next result is useful:

Lemma. Let $(K, d)$ be a compact metric space, and let $(x_n)_{n\geq 1}$ be a sequence in $K$ such that $d(x_{n+1}, x_n) \to 0$. Then the set $E$ of the limit points of $(x_n)_{n\geq 1}$ is connected.
Proof of Lemma. Suppose otherwise $E$ is disconnected. Since $E$ is compact, it is written as a union of two non-empty disjoint compact sets $E_1$ and $E_2$.
Since $\operatorname{dist}(E_1, E_2) > 0$, we can find disjoint open sets $U_1$ and $U_2$ such that $E_i \subseteq U_i$ for each $i$ and $\operatorname{dist}(U_1, U_2) > 0$. Then $(x_n)$ visits both $U_1$ and $U_2$ infinitely often. Also, since $d(x_{n+1}, x_n) \to 0$, infinitely many terms of $(x_n)$ lie outside of $U_1 \cup U_2$. So by the compactness, $(x_n)$ must have a limit point outside $U_1 \cup U_2$, contradicting the choice of $E_i$'s. $\square$

Now we come back to the original question. Since $C(X)$ is separable, the probability measures on $X$ forms a compact metrizable space $\text{Prob}(X)$. Moreover, $\mu_n = \frac{1}{n}\sum_{k=0}^{n-1} \delta_{T^k x}$ satisfies $d(\mu_{n+1}, \mu_n) \to 0$, where $d$ is any "translation-invariant" metric realizing the topology of $\text{Prob}(X)$ (such as the one discussed in this answer). Therefore it follows that the set of limit points of $(\mu_n)$ is connected.
