I'm undergraduated student studying basic topology. While solving the exercises in the Munkres-Topology, one asked me to prove that $\mathbb{R}^*_K$ is not Hausdorff. (section 22, problem 6-(a))
$\mathbb{R}_K$ is a topological space on $\mathbb{R}$ where the basis is given by the form of $(a,b)$ and $(a,b) - K$ ($K = \{ \frac{1}{n} | n \in \mathbb{N} \}$), and $\mathbb{R}^*_K$ is defined by the quotient space obtained from $\mathbb{R}_K$ where the entire set $K$ is considered as a point.
Every solution that I could find on the internet were solving this problem by the following logic: there are no disjoint neighborhoods of $0$ and $K$ in $\mathbb{R}^*_K$, since every open set $(a,b)$ containing $K$ must intersect with $(c,d)$ or $(c,d)-K$ that contains $0$. For instance, $\prod_{i=1}^{\infty} U_i $ can be a neighborhood of $K$ where $U_i = (\dfrac{1}{i} - \epsilon_i , \dfrac{1}{i} + \epsilon_i)$, which (I think that) behavior is totally different from just $(a,b)$.
I can fully understand the statement "every open set $(a,b)$ containing $K$ must intersect with neighborhood containing $0$", but I think that $(a,b)$ just indicates the basis (kind of... I haven't learned any information about basis in the quotient topology) so only dealing about $(a,b)$ cannot fully describe every open set containing $K$.
I'll be very thankful if someone tells me a reason that treating only $(a,b)$ are enough for the solution. Or if it isn't, some guidelines for this problem will be very helpful.
Thank you in advance.
$\Bbb R_K$ was invented to be a Hausdorff space with a closed set $K$ such that $0 \notin K$ but $0$ and $K$ do not have disjoint open neighbourhoods (showing it's not regular). Assume this fact for this exercise.
Now we form the quotient by identifying $K$ to a point (and leaving all points outside $K$ in their own class). The quotient map $q: \Bbb R_K \to X$ (where $X$ is that quotient, so the set of classes $\{\{x\}\mid x \notin K\} \cup \{K\}$) is closed and continuous, $\Bbb R_K$ is Hausdorff but $X$ is not:
Suppose we could separate the points $K$ and $\{0\}$ in $X$ by disjoint open sets $U$ and $V$ respectively. This would imply that $q^{-1}[U]$ is an open neighbourhood of $K$ and $q^{-1}[V]$ of $0$, and disjoint still. But this contradicts the basic fact about $\Bbb R_K$ from the start. So $\{0\}$ and $K$ cannot be separated in $X$ so $X$ is not Hausdorff.