topology basis, euclidean topology Let $k$ be a natural number ($k\in\{0,1,2,...\}$). Let $\mathcal{B}_k$ be the family of subsets of $\mathbb{R}$ consisting of $\mathbb{R}$ and all the open intervals $(a,b)$ which contain at most $k$ integers (when $k=0$, that means: intervals $(a,b)$ that contain no integers). Show that
a. $\mathcal{B}_k$ is a topology basis on $\mathbb{R}$. 
b. for $k\geq 1$, the topology induced by $\mathcal{B}_k$ on $\mathbb{R}$ is the euclidean topology.
Can somebody help me? Thank you very much! 
 A: For part (a.) indeed $\mathcal B_0$ is a basis, but the topology is not the euclidean. Every integer has only one neighborhood, namely $\mathbb R$. 
For part (b.) ($k\ge1$) note that if $x$ is not an integer then $\mathcal B_k$ contains all sets $(x-\varepsilon,x+\varepsilon)$ for every $\varepsilon>0$ with $\varepsilon<d:=\min\{x-\lfloor x\rfloor,\lfloor x\rfloor +1-x\}$.  (Here $\lfloor x\rfloor$ is the floor function, the largest integer smaller than $x$, so $d$ is the distance from $x$ to the nearest integer.) 
On the other hand, if $n$ is an integer, then $\mathcal B_k$ contains all sets $(n-\varepsilon,n+\varepsilon)$ for every $\varepsilon>0$ with $\varepsilon<1$. 
This shows that (for $k\ge1$), the topology induced by the $\mathcal B_k$ is at least as strong as the Euclidean topology. Since every interval $(a,b)$ is open in the Euclidean topology, we also trivially have that the topology induced by $\mathcal B_k$ is weaker than the Euclidean topology, hence the two topologies coincide. 
Finally, concerning $B_1,B_2\in \mathcal{B}_k$, note that if $B_1,B_2$ each contains at most $k$ integers, then so does their intersection 
$B_1\cap B_2$. So you could let $B=B_1\cap B_2$ (as indicated already in the comments). 
A: Let $\tau$ be the Euclidean topology and $\tau'$ be the topology induced by $\mathcal{B}_k$. The collection of all open intervals in $\mathbb{R}$ is a basis for $\tau$. To show $\tau=\tau'$ it suffices to show


*

*For every $x \in \mathbb{R}$ and $B \in \mathcal{B}_k$ containing $x$, there is an open interval $I \subseteq \mathbb{R}$ such that $x \in I \subseteq B$.

*For every $x \in \mathbb{R}$ and $I \subseteq \mathbb{R}$ containing $x$, there is a $B \in \mathcal{B}_k$ such that $x \in \mathcal{B}_k \subseteq I$.


The first bullet point should be immediate. For the second, notice that if $I$ contains $0$ or $1$ integers the result is immediate. Otherwise you need to show that there is a subinterval of $I$ containing $x$ which contains $0$ or $1$ integers. As suggested in the other answer, consider whether $x$ is an integer or not, and in both cases show that there are arbitrarily small intervals centered at $x$ which contain $0$ or $1$ integers. Since these intervals can be made arbitrarily small, you can certainly find one contained in $I$.
