Let $K$ be the set $\{1,1/2,1/3,\cdots,1/n,\cdots\}$, which is the set of reciprocal of all positive integers. The $K$-topology on $\mathbb R$ is defined as generated by the usual open intervals $(a,b)$ and also $(a,b)-K$. For the text below allow me to write $\mathbb R_K$ for the $K$-topology.
The question is: Let $Y$ be the quotient space obtained from $\mathbb R_K$ by collapsing the set $K$ to a point; let $p:\mathbb R_K\to Y$ be the quotient map. Show that $Y$ satisfies the $T_1$ axiom, but is not Hausdorff.
I let $p(1/n)=b$ for all $n\in \mathbb Z_+$, it also map other real numbers into itself. From here I'm understanding that $Y=(\mathbb R\setminus K)\cup\{b\}$, where $b$ can be an element of $K$, or can not.
I can show that $Y$ satisfies $T_1$-axiom. Another problem want me to show that $Y$ is not a Hausdorff space, so I need to find two different point $x_1,x_2$ in $Y$ so that every pair of their respective neighborhood $U_1,U_2$ has nontrivial intersection, right?
My work is as follow: If $b\neq 0$, then we can consider the two points $b,0$ in $Y$. If $U$ is a neighborhood of $0$, then there is $r>0$ such that $(-r,r)\subset U$. If $V$ is a neighborhood of $b$, this means $p^{-1}(V)$ is open in $\mathbb R_K$. But this means $p^{-1}(V)$ contains $K$, so it contains every open set $(1/n-r_n,1/n+r_n)$ for some appropriate $r_n$. Maps this back to $Y$ we see $V$ contains these neighborhoods $(1/n-r_n,1/n+r_n)\setminus\{1/n\}$. Therefore, for $n$ large enough, $1/n$ would smaller than $r$, this makes $U$ and $V$ intersects, then we are done.
However, if $b=0$ then I've no idea how to choose the other point. I think I've messed up some concept about the quotient space.
From: Munkres General Topology, Chapter 22 Question 6(a).
Best Answer
Formally the quotient space $Y$ is actually the set
$$\big\{\{x\}:x\in\Bbb R\setminus K\big\}\cup\{K\}\;;$$
its points are the sets $\{x\}$ for $x\in\Bbb R\setminus K$ and the set $K$. However, in this case there is no real harm in thinking of the quotient as $X=(\Bbb R\setminus K)\cup\{b\}$, where each $x\in\Bbb R\setminus K$ corresponds to the point $\{x\}$ in the quotient, and the point $b$ of $X$ corresponds to the point $K$ of the quotient; you just have to make sure that you get the topology right, which you did when you described a nbhd of $b$. And just as the point $K$ of the quotient is different from each of the points $\{x\}$ for $x\in\Bbb R\setminus K$ — in particular, $K\ne\{0\}$ — so your point $b$ is different from each of the other points of $X$, and in particular, it is not $0$: $b\ne 0$. And as you correctly showed, $0$ and $b$ do not have disjoint open nbhds in $X$, so $X$ is not Hausdorff, and neither is $Y$.