What does it mean for a function of two arguments $f: [0, \infty) \times M \to M$ to be continuous?

Question

Suppose we have a metric space $(M,d)$

Let $f: [0, \infty) \times M \to M$.

What does it mean for $f$ to be continuous?

Here are some suggestions:

1. $f$ is continuous if for every open set $V \subseteq M$ , $f^{-1}(V)$ is open in the product topolgy of $[0 , \infty) \times M$, where the product topology is the topology generated by the natural topology on $[0, \infty)$ and the metric topology on $M$.
2. $f$ is continuous if for all sequences $(t_k, x_k) \to (t,x) $, $f(t_k,x_k)$ converges to $f(t,x) \in M$
3. $f$ is continuous if for all $(t,x) \in [0 \times \infty) \times M$, and for all $\epsilon >0$, there exists a $\delta >0$, such that if $(t',x') \in [0, \infty) \times M$ whenever $d((t,x), (t',x')) < \delta \implies d(f(t,x), f(t',x')) < \epsilon$

Can someone check my definition? And are these conditions equivalent?

Note: the last definition is messed up, I just noticed $d((t,x), (t',x'))$ doesn't make sense.

Answer 1

All 3 are equivalent, provided you take a correct metric on $[0,\infty) \times M$. We could take $D((t_1,x_1), (t_2,x_2)) = \max(|t_1 - t_2|, d(x_1,x_2))$ and that metric $D$ induces the product topology on $[0,\infty) \times M$.

So the reformulated version of 3. would be:

$f$ is continuous if for all $(t,x) \in [0 \times \infty) \times M$, and for all $\epsilon >0$, there exists a $\delta >0$, such that if $(t',x') \in [0, \infty) \times M$ whenever $(|t-t'| < \delta \land d(x,x') < \delta) \implies d(f(t,x), f(t',x')) < \epsilon$

because $\max(a,b) < \delta$ iff $a < \delta \land b < \delta$. We could also have used a Euclidean combination like $D((t_1,x_1), (t_2,x_2)) = \sqrt{(t_1-t_2)^2 + d(x_1,x_2)^2}$ or $D((t_1,x_1), (t_2,x_2))= |t_1 + t_2| + d(x_1,x_2)$, but the $\max$-variant gives the "nicest" reformulation IMHO.

You're basically restating 1. in two other guises: epsilon-delta in 3 and sequentially in 2. So I'd say 1. is the correct interpretation of the question in the title (taking the product topology and the default topology on $[0,\infty)$) and by standard facts 2. and 3. are equivalent in the metric setting that you're in.