Quotient topology of $[0,1]$

Question

Is it possible to define an equivalence relation $\sim$ on $[0,1]$ in such a way that $[0,1]/\sim$ is not second countable?

Answer 1

For $x,y\in[0,1]$ let $x\sim y$ iff $x=y$ or $x,y\in\Bbb Q$, and let $X=[0,1]/\sim$; $X$ is the result of identifying $\Bbb Q\cap[0,1]$ to a point. Let $p$ be the point of $X$ corresponding to the $\sim$-equivalence class $\Bbb Q$, and let $q:[0,1]\to X$ be the quotient map. A set $U\subseteq X$ is an open nbhd of $p$ iff $q^{-1}[U]$ is an open nbhd of $\Bbb Q\cap[0,1]$ in $[0,1]$, which is the case if and only if $[0,1]\setminus q^{-1}[U]$ is a closed subset of $[0,1]\setminus\Bbb Q$.

If $X$ has a countable local base at $p$, there is a countable family $\mathscr{F}=\{F_n:n\in\Bbb N\}$ of closed subsets of $[0,1]\setminus\Bbb Q$ such that if $H$ is any closed subset of $[0,1]\setminus\Bbb Q$, there is an $n\in\Bbb N$ such that $H\subseteq F_n$. (Just take the complements in $[0,1]$ of the sets $q^{-1}[U]$ with $U$ in the countable local base at $p$.) Thus, to show that $X$ has no countable local base at $p$, we need only show that if $\mathscr{F}=\{F_n:n\in\Bbb N\}$ is any countable family of closed subsets of $[0,1]\setminus\Bbb Q$, there is a closed subset $H$ of $[0,1]\setminus\Bbb Q$ such that $H\setminus F_n\ne\varnothing$ for each $n\in\Bbb N$.

Each $F_n$ is closed in $[0,1]$ and disjoint from $\Bbb Q$, so each $F_n$ is a closed, nowhere dense subset of $[0,1]$. $[0,1]$ is a compact Hausdorff space, so by the Baire category theorem it is not the union of countably many closed, nowhere dense subsets. We can therefore choose an irrational $x\in[0,1]\setminus\bigcup\mathscr{F}$ and let $H=\{x\}$; clearly $H\setminus F_n=H\ne\varnothing$ for each $n\in\Bbb N$, as desired.

Answer 2

Yes.

Consider the equivalence relation $\sim$ obtained by identifying all points of the form $\frac{1}{n}$ (and being just equality on the rest of the unit interval). Denoting by $*$ the equivalence class of the reciprocals of the natural numbers in $[0,1]/\sim$, I claim that there is no countable basis at $*$. Note that open neighbourhoods of $*$ essentially look like $$\{ * \} \cup \bigcup_{n=1}^\infty \left( ( \tfrac{1}{n} - \epsilon_n , \tfrac 1n ) \cup ( \tfrac 1n , \tfrac 1n + \epsilon_n ) \right)$$ where $0 < \epsilon_n \leq \frac{1}{n(n+1)}$. If $\{ V_m \}_{m=1}^\infty$ is any countable collection of such open neighbourhoods, say $$V_m = \{ * \} \cup \bigcup_{n=1}^\infty \left( ( \tfrac{1}{n} - \epsilon_{m,n} , \tfrac 1n ) \cup ( \tfrac 1n , \tfrac 1n + \epsilon_{m,n} ) \right)$$ for each $m$, consider $$U = \{ * \} \cup \bigcup_{n=1}^\infty \left( ( \tfrac{1}{n} - \tfrac{\epsilon_{n,n}}2 , \tfrac 1n ) \cup ( \tfrac 1n , \tfrac 1n + \tfrac{\epsilon_{n,n}}2 ) \right).$$ This is clearly an open neighbourhood of $*$, but $V_m \not\subseteq U$ for all $m$ (meaning that the $\{ V_m \}_{m=1}^\infty$ is not a local base at $*$). (Note that this shows that $[0,1]/\sim$ is not first-countable.)