Proof that the quotient space of $\mathbb{R}^2/L$ where $L$ is a line passing through the origin is not first countable

Question

Let $L$ be a line in the plane $\Bbb R^2$ passing through origin. How would you prove that the quotient space of $\frac{\mathbb{\Bbb R}^2}{\sim}$ where $\sim$ is equivalence relation defined by $a\sim b$ iff $a=b$ or $a,b\in L$.

How do you prove that this space is not first countable (not every point has a countable fundamental system of neighborhoods)? My idea was to consider the point $L$ collapses to and the fact that every one of it's open neighborhoods is an open set of $\mathbb{R}^2$ containing $L$. Then I would need to show that for every countable set $S$ of open sets containing $L$ I can always find an open set $A$ that is not the superset of any of the elements of $S$. How do I show this?

Answer 1

Without loss of generality, $L=\mathbb R \times \{0\}.$ Suppose $\{U_n\}$ is a countable neighborhood base at $[0]=\mathbb R\times \{0\}.$ Since $U_n$ is open in the quotient, $\pi^{-1}(U_n)$ is open in $\mathbb R^2$ and contains $[0]=\mathbb R\times \{0\}$ so for each integer $n$, there is a point $x_n\in \mathbb R^+$ such that $(n,x_n)\in U_n.$

But now, if we define $U=\mathbb R^2\setminus \{(n,x_n):n\in \mathbb N\}$, then $\pi(U)=[0]\bigcup (R^2\setminus \{(n,x_n):n\in \mathbb N\})$ and so $\pi^{-1}(\pi(U))=R^2\setminus \{(n,x_n):n\in \mathbb N\},$ which is open in $\mathbb R^2$ and this means that $\pi(U)$ is open in the quotient. And yet it is not contained in any $U_n$, so $\{U_n\}$ cannot be a neighborhood base.

Answer 2

Without loss of generality assume $L$ is the $X$-axis. Cosinder the set $$B=\bigg\{\bigg[\bigg(x,\frac{1}{n}\bigg)\bigg]:x\in\Bbb R,n\in \Bbb N\bigg\}\subseteq\frac{\Bbb R^2}{L}.$$ Then, $$[l]\in \overline B,\text{ for }l\in L.$$ We can show that, there is no sequence from $B$ converging to $[l]$ : Choose any sequence $\big\{\big[\big(x_k,\frac{1}{n_k}\big)\big]\big\}_{k\in \Bbb N}$ of points from $B$. Consider $L\subseteq U\subseteq_{\text{open}}\Bbb R^2$ such that, $\big(x_k,\frac{1}{n_k}\big)\not\in U$ for all $k$. Then $q(U)$ is an open set containing $[l],l\in L$ but, $\big[\big(x_k,\frac{1}{n_k}\big)\big]\not\in q(U)$. Here $q:\Bbb R^2\to\frac{\Bbb R^2}{L}$ defined by $(a,b)\mapsto[(a,b)],(a,b)\in\Bbb R^2$ is the quotient map.

So $\frac{\Bbb R^2}{L}$ is not first countable.

Let $X$ be first countable topological space and $A\subseteq X$. Then, $x\in \overline A$ implies that, there is a sequence $\{a_n\}\subseteq A$ converging to $x$. To prove this, choose a local countable base $\{U_n\}$ at $x$ and fix points $a_n\in U_n\cap A$. This is possible as $x\in \overline A$.