Equivalence of two definitions of continuous-time dynamical system

Question

I'm new to dynamical systems. I've found out that there are at least two possible approaches to defining continuous-time dynamical systems but they are equivalent: One is a tuple $(\mathbb{R},X,\phi)$(of course it can be more general than that but I've restricted our discussion to this) such that $X$ is a metric space and $\phi$ is a continuous function from $\mathbb{R} \times X$ to $X$ that satisfy two properties(in other words $\phi$ is a left group action of $\mathbb{R}$ on $X$), and the other is a family of functions $(\phi_t)_{t \in \mathbb{R}}$ from $X$ to $X$ such that $\phi_0 = $ id$_X$ and $\phi_{t + s} = \phi_t \circ \phi_s$. I can prove that the first one induces a family of functions $(\phi^t)_{t \in \mathbb{R}}$ satisfying the given properties. So one direction is OK. And I can prove that the second one induces a function $\phi$ satisfying all the stated properties other than one: Continuity. So the question is how to prove that $\phi$ is continuous. I've defined $\phi$ to map $(t,x)$ to $\phi_t(x)$. Thanks.

Answer 1

This is false in general, if you use rhe second definition you would need to explicitly require continuity, here is a counter example:

Let $R$ be the real numbers with the discrete topology (0,1 metric), and define $\phi_t : R \rightarrow R$ by $\phi_t(r) = r+t$. All these functions are continuous since $R$ has the discrete topology. These induce the function $$\phi : \mathbb{R} \times R \rightarrow R$$ $$\phi(t,r) = \phi_t(r)$$ Now, note that $\phi$ is not continuous, since it carries connected components into disconnected ones: to see this just observe that $\phi$ restricted to $\mathbb{R}\times \{0\}$ is surjective.