Cover $(0, +\infty )$ by open sets Cover $(0, +\infty)$ by open sets $U_\alpha$ such that for any $\epsilon &gt; 0$ there are points $x, y \in (0, +\infty)$ with $|x-y|&lt;\epsilon$, not both belonging to the same $U_\alpha$


The distance beteen $x$ and $y$ is $\epsilon$ so would it work if we took our open setss $U_\alpha=B_{\frac{\epsilon}{2}}(\alpha)=\{x \in \mathbb{R}| d(x, \alpha) &lt;\frac{\epsilon}{2} \}$

So $|x-y|&lt;\epsilon$ but we can take our open ball as having a radius$\frac{\epsilon}{2} $

I feel like I may have missed something - is my answer correct?
$$\mathscr U:=\{B_{1/n}(n):n\in\mathbb N\}\cup \{(n,n+1):n\in\omega\}$$ is an open cover of $(0,\infty)$. 

Now let $\epsilon&gt;0$. There exists $n\in\omega$ such that $1/n&lt;\epsilon/3$. Then $n-\epsilon/3$ and $n+\epsilon/3$ are distance less than $\epsilon$ apart but are not in any single member of $\mathscr U$.

I approached the problem by thinking about simple open covers - open covers consisting of intervals. I noted that such open cover cannot have a finite subcover if we want your property. So then I thought about my favorite such open cover of $\mathbb R$, which is all the sets $(n,n+1)$ and $(n−1/2,n+1/2)$ (pictured in my head). Then I just thought about shrinking the $(n−1/2,n+1/2)$  intervals so that I can always get close-together $x$ and $y$ at the "ends" of one of them.
Your definition of $U_{\alpha}$ is not clear, is $\alpha$ a point or is it an index? 

Besides that, you want to be able to find these two points for every $\epsilon$.

If I understand correctly, what you need is a family of sets such that you can always find two points very close to each other such that they correspond to different $U_{\alpha}$'s. You can go overkill and try $U_{x,\epsilon}:=B_{\epsilon}(x)\cap (0,+\infty)=\{y\in (0,+\infty)\mid d(x,y)&lt;\epsilon\}$, for every point $x\in (0,+\infty)$ and every $\epsilon &gt;0$.

So now let $\epsilon &gt;0 $, take any $x\in (0,+\infty)$ and the neighborhood $U_{x,\frac{\epsilon}{4}}$ and $y:=x+\frac{\epsilon}{2}$ and the neighborhood $U_{y,\frac{\epsilon}{4}}$, you know that $d(x,y)=\frac{\epsilon}{2} &lt; \epsilon $ and that $U_{x,\frac{\epsilon}{2}}\cap U_{y,\frac{\epsilon}{4}}=\emptyset$