As the title explains, I'm trying to solve a problem which asks me to find the closure of $(0,1)$ with respect to the topology $\mathcal{T}$ which is the family of subsets $U$ of $\mathbb{R}$ with the property that for every $ x\in U$, there exists $ε>0$ such that $(x-ε, x]\subseteq U$.
I've already proven that $\mathcal{T}$ actually is a topology, but I'm struggling to see what the closure of $(0,1)$ is, much less how to prove it.
I'd appreciate any help you could offer.
Closure of $(0,1)$ is $(0,1]$. For $1$, every $(1-\epsilon,1]\cap(0,1)\neq \emptyset$, so $1$ is in the closure of $(0,1)$. F0r $a\leq 0$, take any $\epsilon $, then $(a-\epsilon,a]\cap(0,1)= \emptyset$. So $a$ is not in the closure of $(0,1)$. For any $a>1$, say $\epsilon=\frac{a-1}{2}$, then $(a-\epsilon,a]\cap(0,1)= \emptyset$.