[proofplan] We prove the two implications separately. A dense forward orbit implies the open-set hitting condition once we observe that, in a space with no isolated points, every tail of a dense orbit is still dense. Conversely, the open-set hitting condition makes the sets of points whose orbits meet a fixed basis element open and dense. Since a non-empty compact [metric space](/page/Metric%20Space) is a Baire space and has a countable basis, the intersection of these open dense sets contains a point whose orbit meets every basis element. [/proofplan]

[step:Show that every tail of a dense orbit is dense]Assume that $x\in X$ has dense forward orbit. Fix $N\in\mathbb{N}_0$ and define the $N$-th orbit tail \begin{align*} \mathcal{O}_{f,N}^+(x):=\{f^n(x):n\in\mathbb{N}_0\text{ and }n\geq N\}. \end{align*} We prove that $\mathcal{O}_{f,N}^+(x)$ is dense in $X$. Let $W\subset X$ be non-empty and open. Define the finite set \begin{align*} F_N:=\{f^0(x),f^1(x),\ldots,f^{N-1}(x)\}, \end{align*} with $F_0:=\varnothing$. Since $X$ has no isolated points, every finite subset of $X$ has empty interior. Hence $W\setminus F_N$ is a non-empty open subset of $X$. The density of $\mathcal{O}_f^+(x)$ gives some $m\in\mathbb{N}_0$ such that $f^m(x)\in W\setminus F_N$. If $m<N$, then $f^m(x)\in F_N$, a contradiction. Thus $m\geq N$, and $f^m(x)\in W\cap \mathcal{O}_{f,N}^+(x)$. Since every non-empty open $W$ meets $\mathcal{O}_{f,N}^+(x)$, the tail is dense.[/step]

[guided]The only point that needs care is that deleting finitely many early orbit points should not destroy the ability to hit an arbitrary [open set](/page/Open%20Set). Fix $N\in\mathbb{N}_0$ and put \begin{align*} \mathcal{O}_{f,N}^+(x):=\{f^n(x):n\in\mathbb{N}_0\text{ and }n\geq N\}. \end{align*} Let $W\subset X$ be non-empty and open. The early part of the orbit is the finite set \begin{align*} F_N:=\{f^0(x),f^1(x),\ldots,f^{N-1}(x)\}, \end{align*} where $F_0=\varnothing$. Because $X$ has no isolated points, no finite subset of $X$ contains a non-empty open set; equivalently, $F_N$ has empty interior. Therefore $W\setminus F_N$ is still a non-empty open set. Since the full forward orbit $\mathcal{O}_f^+(x)$ is dense, it meets this smaller open set. Choose $m\in\mathbb{N}_0$ such that $f^m(x)\in W\setminus F_N$. This point cannot come from one of the first $N$ iterates, because those iterates all lie in $F_N$. Hence $m\geq N$. We have found a point $f^m(x)$ with $m\geq N$ and $f^m(x)\in W$, so $W$ meets the $N$-th tail. Since $W$ was arbitrary, $\mathcal{O}_{f,N}^+(x)$ is dense in $X$.[/guided]

[step:Derive open-set transitivity from a dense orbit] Assume that $x\in X$ has dense forward orbit. Let $U,V\subset X$ be non-empty and open. By density, choose $m\in\mathbb{N}_0$ such that $f^m(x)\in U$. By the previous step applied with $N=m+1$, the tail $\mathcal{O}_{f,m+1}^+(x)$ is dense, so there exists $k\in\mathbb{N}_0$ with $k\geq m+1$ and $f^k(x)\in V$. Set $n:=k-m\in\mathbb{N}$. Since $f^m(x)\in U$ and $f^n(f^m(x))=f^k(x)\in V$, we have \begin{align*} f^n(U)\cap V\ne\varnothing. \end{align*} Thus the open-set hitting condition holds. [/step]

[step:Build open dense hitting sets from open-set transitivity] Assume now that for every pair of non-empty open sets $U,V\subset X$, there exists $n\in\mathbb{N}_0$ such that $f^n(U)\cap V\ne\varnothing$. Since $X$ is a compact metric space, it is second countable. Choose a countable basis $(B_j)_{j=1}^{\infty}$ for the topology of $X$ consisting of non-empty open sets. For each $j\in\mathbb{N}$, define \begin{align*} G_j:=\bigcup_{n=0}^{\infty} f^{-n}(B_j). \end{align*} Each set $G_j$ is open because $f^n:X\to X$ is continuous for every $n\in\mathbb{N}_0$. We prove that $G_j$ is dense. Let $U\subset X$ be non-empty and open. Applying the open-set hitting condition to $U$ and $B_j$, choose $n\in\mathbb{N}_0$ such that $f^n(U)\cap B_j\ne\varnothing$. Then there exists $u\in U$ with $f^n(u)\in B_j$, so $u\in U\cap f^{-n}(B_j)\subset U\cap G_j$. Thus every non-empty open $U$ meets $G_j$, and $G_j$ is dense in $X$. [/step]

[step:Apply Baire category to obtain a point with dense orbit] For each $j\in\mathbb{N}$, the previous step shows that $G_j$ is open and dense in $X$. Since $X$ is a non-empty compact metric space, the [Baire Category Theorem](/theorems/630) applies to $X$. Therefore \begin{align*} G:=\bigcap_{j=1}^{\infty}G_j \end{align*} is dense in $X$, and in particular $G\ne\varnothing$. Choose $x\in G$. For every $j\in\mathbb{N}$, the membership $x\in G_j$ means that there exists $n_j\in\mathbb{N}_0$ with $f^{n_j}(x)\in B_j$. Hence the forward orbit of $x$ meets every basis element $B_j$. Let $W\subset X$ be non-empty and open. Since $(B_j)_{j=1}^{\infty}$ is a basis consisting of non-empty open sets, there exists $j\in\mathbb{N}$ such that $B_j\subset W$. The orbit of $x$ meets $B_j$, so it also meets $W$. Hence every non-empty open subset of $X$ meets $\mathcal{O}_f^+(x)$, which is exactly the statement that $\mathcal{O}_f^+(x)$ is dense in $X$. [/step]