Almost every $x$ is a cluster point of its own trajectory. The following problem appears in [1]:
2.3.2(a) Let $(X, d)$ be a compact metric space and let $T:X \rightarrow X$ be a continuous map. Suppose that $\mu$ is a $T$-invariant probability measure defined on the Borel subsets of $X$. Prove that for $\mu$-almost every $x \in X$ there is a sequence $n_k \rightarrow \infty$ with $T^{n_k}(x) \rightarrow x$ as $k \rightarrow \infty$.
(b) Prove that the same conclusion holds under the assumption that $X$ is a metric space, $T:X \rightarrow X$ is Borel measurable, and $\mu$ is a $T$-invariant probability measure.
This question appears in the section on Poincaré recurrence. Intuitively, it makes sense and should obviously be true. However, I am having difficulty writing a nice proof, mainly from the fact that $x$ is a specific point in its own neighborhood, and I cannot be certain when changing neighborhoods that the collection of almost every point returning is the same.
My intuition is telling me that I might be able to use the Lebesgue number of an open cover of $\varepsilon$-balls to simplify the argument. However, part (b) does not assume compactness (or even completeness).
What I want to say is:
Let $A_k = \{x \in X : \exists n_k \in \mathbb{N}, T^{n_k}(x) \in B_k(x)\}$, where $B_k(x) = \{y \in X : d(x, y) < 1/k\}$, for all $k \in \mathbb{N}$. By Poincaré recurrence, $\mu(A_k) = \mu(X) < \infty$. Let $B = \cap_{j=1}^\infty A_j$. Then, $B$ is precisely $\{x \in X : \exists n_k \rightarrow \infty, T^{n_k}(x) \rightarrow x \mathrm{\ as\ } k \rightarrow \infty\}$, and $\mu(B) = \mu(X)$.
Question: How can I fix this to make a correct (and not ugly) proof?
[1] Einsiedler, M., & Ward, T. (2011). Ergodic theory: With a view towards number theory. London: Springer-Verlag.
 A: In case (a), let $\mathscr{B} = (B_n)_{n\geq 1}$ be a countable basis of open balls (which exists by compactness).
Let's call a point $x$ recurrent if there exists a subsequence $(n_k)$ with $T^{n_k}x\rightarrow x$, and all other points non-recurrent. If $x$ is a non-recurrent point, then that means there exists $\epsilon > 0$ such that eventually (meaning for all $k\geq$ some $N$)
$$
d(T^k x,x)\geq \epsilon.
$$
Choose $B_n$ such that $x \in B_n \subset B(x,\epsilon)$. The Poincare recurrence theorem applied to $B_n$ implies that there is some $C_n\subset B_n$ with full measure such that every point in $C_n$ visits $B_n$ infinitely often. Then our point $x$ above obviously can't belong to $C_n$, since $B_n\subset B(x,\epsilon)$. We have shown that if $x$ is non-recurrent, then
$$
x \in \bigcup\limits_{n=1}^\infty B_n\setminus C_n
$$
where $C_n$ is the full-measure set obtained by applying Poincare recurrence to $B_n$. This is a zero measure set since each $B_n\setminus C_n$ is zero measure.
Don't know yet about part (b) but I've fav'ed this for when someone else answers.
