A question about *continuity* of group actions

Question

I'm taking back some definitions and concepts from old algebra/topology studies and I'm trying to show the following simple thing. Given a group action

\begin{align}&\mathbb{Z} \times \mathbb{R} \rightarrow \mathbb{R}\\ &(n,x) \longmapsto n \cdot x := x+n\end{align}

i.e. the left-translation of a real number by an integer $n$. So I want to show that this map is continuous with respect to the natural topology in $\mathbb{R}$ and product topology $\mathbb{Z} \times \mathbb{R}$.

Since for every $n \in \mathbb{Z}$ we can define a map

$$\varphi_n : \mathbb{R} \rightarrow \mathbb{R}$$

such that $\varphi_n(x) := x+n, \, \, \varphi^{-1}_n(x) = x-n$, then considering an open set $A = B_\varepsilon(x+n)$ how to show rigorously that $\varphi^{-1}(A)$ is open?

Is this a good start?

EDIT:

Trying to follow hints:

1. If $U \subset \mathbb{R}$ open then $\forall u \in U \, \, \exists \varepsilon >0 \, : \, B_{\varepsilon}(u) \subseteq U$. But then I would have to show that $B_{\varepsilon'}(u-x) \subseteq U-x$ for some new $\varepsilon'$?

2)Is this something that I should prove? Even if we have product topology I would say that $\mathbb{Z} \subset \mathbb{R}$ then it can be considered the subspace topology where open sets of $\mathbb{Z}$ are intersections of sets in $\mathbb{Z}$ with open sets of $\mathbb{R}$ (exactly what suggested). But how does this link with the product topology?

Answer 1

Proof sketch for hint 1: Suppose that $U \subset \mathbb{R}$ is open. Define the set $U - x_0$ as $\{u - x_0 : u \in U\}$. Suppose $y \in U - x_0$. Then $y + x_0 \in U$, so there's some $\epsilon > 0$ such that $U$ contains the entire interval $(y + x_0 - \epsilon, y + x_0 + \epsilon)$. This means that $U - x_0$ contains the interval $(y - \epsilon, y + \epsilon)$.

Proof sketch for hint 2: Remember that the product topology on $X \times Y$ is generated by open sets of the form $U \times V$, where $U \subseteq X$ and $V \subseteq Y$ are open. The topology on $\mathbb{Z}$ is the discrete topology, so the singleton sets $\{n\}$ are open, and so are sets $\{n\} \times U$ where $U \subset \mathbb{R}$ is open. Any infinite union of these sets is also open.

Finally, $\varphi^{-1}(A)$ (it's not hard to show) is a union of sets of the form $\{n\} \times (A - n)$ for every integer $n \in \mathbb{Z}$.