Let $X$ be a compact metric space and $T:X\rightarrow X$ be a continuous map. Let $\mu$ be a $T$-invariant Borel measure such that $\mu(U)>0$ for each open set $U \subset X$.
I'm trying to show that \begin{equation*} \mu \{ x \in X: \text{the set} \{T^n(x):n \in \mathbb{N} \} \text{ is dense in } X \} = 1. \end{equation*}
Apparently the solution involves finding a basis for the topology, but unfortunately I haven't studied topology so I'm not sure how this helps. Ergodicity will also probably be useful.
Any help would be appreciated!
Edit: Sorry for the very late response, turns out the question was missing the fact that $T$ is ergodic with respect to $\mu$.
A couple things: one, you seem to be implicitly assuming that $\mu$ is a probability measure, as you seem to be wanting to prove that almost every point has dense orbit. Two, as the comments have shown, this doesn't seem true if you only assume $T:X\to X$ continuous. However, it does hold if you assume that $T$ is ergodic with respect to $\mu$.
For that, recall the Birkhoff Ergodic Theorem. You need only prove that the set $\{T^n(x)\}_{n=1}^{\infty}$ intersects each nonempty open set $U$ to prove density. By the theorem, $\frac{1}{n} \sum_{i=1}^n \chi_U(T^i x)\to \int_X \chi_U d\mu=\mu(U)>0$ for $\mu$-a.e. $x$, so in fact, infinitely many of the iterates must intersect $U$ for almost every $x$. If you assume unique ergodicity, you would in fact have every orbit dense in $X$, but that requires a tiny bit more work.