A topology for the real line other than the standard one I am reading the book Introduction to Topology by V. A. Vassiliev. He defines basis as follows:

A basis of a topology $\mathcal{T} = \{ V_{\alpha} \}_{\alpha \in
 \Gamma } $ is a subset $\{ W_{\lambda} \}_{\lambda \in \Gamma} \subset
 \mathcal{T}  $ such that each open set can be represented as the union
   of a (probably infinite) family of sets from $W_{\lambda}$.

He then gives as an example of a topology: Take $\mathbb{R}$. For the open sets we take those sets that are open in the usual sense, and periodic with period one:
$$ t \in U \iff t + 1 \in U $$
for each open $U$.
MY question is: How is this a topology? I am having hard time trying to see how are the open sets in this topology. Also, what is the basis for this topology?
thanks
 A: An example of an open set is $$\bigcup_{n\in\Bbb Z}\left(n,n+\frac12\right)\;;$$ it’s certainly open in the usual topology, and it’s also periodic with period $1$. You can get a base for the new topology as follows.
Let $\mathscr{B}$ be any base for the usual topology; for instance, $\mathscr{B}$ could be the entire topology, or the set of open intervals, or the set of open intervals with rational endpoints. For each $B\in\mathscr{B}$ let
$$B+\Bbb Z=\{x+n:x\in B\text{ and }n\in\Bbb Z\}\;.$$
If we write $B+n$ for $\{x+n:x\in B\}$, then
$$B+\Bbb Z=\bigcup_{n\in\Bbb Z}(B+n)=\ldots\cup(B-2)\cup(B-1)\cup B\cup(B+1)\cup(B+2)\cup\ldots\;;$$
clearly $B+\Bbb Z$ is open in the usual topology and periodic with period $1$. I claim that $\{B+\Bbb Z:B\in\mathscr{B}\}$ is a base for the new topology.
To see this, let $U$ be open in the new topology. Then $U$ is open in the usual topology, so there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$, and I claim that in fact $$U=\bigcup\{B+\Bbb Z:B\in\mathscr{B}_U\}\;;$$ if you can show this, you’ve shown that $\{B+\Bbb Z:B\in\mathscr{B}\}$ is a base for the new topology. For now I’ll leave it to you; it’s not too hard. If you get stuck, leave a comment.
A: Think of the usual topology of $(0,1)$, a subset $A$ of $(0,1)$ is in this topology if and only if $\cup_{n=-\infty}^{+\infty}(A+n)$ is in the topology that you describe
A: Examples of open sets
$$\Bbb R\setminus\Bbb Z$$
$$\cdots\cup(-2,-3/2)\cup(-1,-1/2))\cup(0,1/2)\cup(1,3/2)\cup\cdots$$
$$\cdots$$
