For the open sets of $\mathbb{R}$ take those sets that are open in the usual sense and periodic with period $1$ ( $t \in U \iff t+1 \in U \; , \forall U \; \; open $) PROVE this is a topology on R
attempt:
I am having some trouble proving this.Trivially $\varnothing $ satisfies this and $\mathbb{R}$ as well.
Now, if we consider a colletion $(U_{\alpha})$ of open sets, then take $x \in \bigcup U_{\alpha}$
then $x+1 \in \bigcup U_{\alpha}$ it follows that $x+1\in U_{\alpha}$ for some $\alpha$ so the union is in the collection
finally, take $U_1$ and $U_2$ and let $x \in U_1$ and $x \in U_2$ so$x+1 \in U_1$ and $x+1 \in U_2$ and so $x + 1 \in U_1 \cap U_2$
so $U_1 \cap U_2$ is in the collection
is this enough to prove this collection is a topology on $\mathbb{R}$?
No, the open sets have the extra condition
$$\forall x \in U: x+1 \in U\tag{1}$$
which is voidly satisfied for $\emptyset$ and trivial for $U=X$.
If $U_i, i \in I$ is a collection of periodic Euclidean open sets, $U:=\bigcup_i U_i$ is Euclidean open (we know that the Euclidean topology is a topology) and to check it is still periodic, take any $x \in U$. Then $x \in U_i$ for some $i$, and as $U_i$ is periodic, $x+1 \in U_i$ and so $x+1 \in U$. So $U$ is periodic (it satisfies $(1)$).
The finite intersection you did OK.