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The convention intended by this question is existence of a dense forward orbit, also called point transitivity. On a nonempty compact metric space, this means that some $x\in X$ satisfiesEquivalently, that forward orbit visits every nonempty open set.

The other common definition of topological transitivity uses intersections of iterated open sets, as in part (b). The distinction matters on spaces with isolated points: parts (b) and (c) ask precisely how the two conventions are related. We use point transitivity for the dense-orbit convention throughout this solution. As usual for this definition, the state space is assumed nonempty; on an empty space the open-set condition is vacuous and there is no point with a forward orbit.

Let $x$ have a dense forward orbit $O=\{f^j(x):j\ge 0\}$, and suppose that $f$ is surjective. Every tail $O_k=\{f^j(x):j\ge k\}$ is then dense. Indeed, continuity givesand the left side equals $f^k(X)=X$ by surjectivity.

Given nonempty open sets $U,V$, choose $k\ge 0$ with $f^k(x)\in U$. The dense tail $O_{k+1}$ supplies $j\ge k+1$ with $f^j(x)\in V$. Taking $n=j-k\ge 1$ gives $f^j(x)\in f^n(U)\cap V$. This proves the required positive-time intersection property.

The result can fail without surjectivity. Take the compact metric spacewith the metric induced from the real line. All nonzero points are isolated, so continuity there is automatic; at zero, $f(1/j)\to 0=f(0)$ proves continuity. The forward orbit of $1$ is $\{1,1/2,1/3,\dots \}$, whose closure is $X$, so this is point transitivity. But $f(X)=X\setminus \{1\}$. The singletons $U=\{1/2\}$ and $V=\{1\}$ are nonempty open sets, and for every $n\ge 1$,The isolated initial point can be visited only at the start of the dense forward orbit; deleting that initial segment destroys density.