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! 

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 $I$.
