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My question is whether or not there exist any quotient spaces of the ordered square $[0,1]^2$ such that it is both Hausdorff and is NOT first-countable.

Attempts at the Solution:

  • My first attempt was to take $[0,1]^2/\Delta$, where $\Delta$ is the diagonal of $[0,1]^2$ (i.e. collapse the diagonal to a point and keep everything else fixed). while this is indeed not first-countable (see Quotients of the Ordered Square) it is clearly not Hausdorff because it is not even $T_0$: the one point set that contains the point $d$ in $[0,1]^2/\Delta$ that is the equivalence class of $\Delta$ is closed iff $\Delta$ is closed in $[0,1]^2$, which it is not. Thus, it isn't even $T_1$, meaning that it certainly cannot be Hausdorff. This is indicative of the fact that it is a necessary condition for the pre-image of the equivalence classes of $[0,1]^2/\sim$ to be closed in $[0,1]^2$.

  • My second attempt was to force it to be $T_1$ be taking $[0,1]^2/\overline{\Delta}$, but this is now first-countable because the only neighborhoods of the equivalence class of $\overline{\Delta}$ (which are open sets on $[0,1]^2$ that contain $\overline{\Delta}$) is the whole quotient space, and this is clearly a countable base about the point. EDIT: these statements are not true; removing any point not in $\overline{\Delta}$ from $[0,1]^2$ gives an neighborhood.

  • My third attempt was to take the quotient space $[0,1]/\sim$ where $(x,y)\sim(w,z)$ iff $x=w$. The elements of this quotient space are the closed vertical intervals of $[0,1]^2$, and this ends up being homeomorphic to $[0,1]$ as a subspace of $\mathbb{R}$. In particular, it is Hausdorff but first-countable.

  • My final attempt was to take the quotient space similar to the first, but this time only take the subset $\{(1/n,1/n)\mid n>0\}$ and closing it by adding the point $(0,1)$. I feel fairly certain that this is Hausdorff, but whether it is first-countable or not I couldn't quite figure out. My motivation for this quotient space was the fact that the quotient space $\mathbb{R}/\mathbb{N}$ is not first-countable.

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For convenience denote the lexicographically ordered square by $X$.

Your first example, $X/\Delta$, is $T_0$; it just isn’t $T_1$ (so of course it isn’t Hausdorff, either). Remember, $T_0$ doesn’t imply that singletons are closed.


Your second example actually does what you want. Let $D=\operatorname{cl}_X\Delta$, and let $p_D$ be the point of $X/D$ corresponding to $D$. It’s not true that the only open nbhd of $p_D$ is $X/D$: if $p=\left\langle\frac12,\frac34\right\rangle$, then $X\setminus\{p\}$ is a saturated open nbhd of $D$ in $X$, so its image under the quotient map $\pi:X\to X/D$ is an open nbhd of $p_D$ not containing $\pi(p)$.

Suppose that $U$ is an open nbhd of $D$ in $X$, and let $F=X\setminus U$. For $x\in[0,1]$ let $$F_x=F\cap\big(\{x\}\times[0,1]\big)\;;$$ each $F_x$ is closed in $X$. Let $A=\{x\in[0,1]:F_x\ne\varnothing\}$. If $A$ is infinite, it has a cluster point $u\in[0,1]$, and either $\langle u,0\rangle$ or $\langle u,1\rangle$ is a cluster point of $F$ and hence in $F\cap D=\varnothing$, which is absurd. Thus, $A$ is finite.

For $x\in[0,1]$ let $\mathscr{C}_x$ be the set of closed subsets of $(0,1)\setminus\{x\}$. For each $x\in[0,1]$ and $C\in\mathscr{C}_x$ let $$U(x,C)=X\setminus\big(\{x\}\times C\big)\;;$$ $U(x,C)$ is an open nbhd of $D$ in $X$. Let $\mathscr{S}=\{U(x,C):x\in[0,1]\text{ and }C\in\mathscr{C}_x\}$, and let $$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{S}\text{ is finite}\right\}\;;$$ it follows from the last paragraph that $\mathscr{B}$ is a nbhd base for the set $D$, and since the members of $\mathscr{B}$ are saturated, $\mathscr{B}'=\{\pi[B]:B\in\mathscr{B}\}$ is a nbhd base at $p_D$ in $X/D$. Moreover, no countable subfamily of $\mathscr{B}'$ is a base at $p_D$: for any countable $\mathscr{B}_0\subseteq\mathscr{B}$, there is an $x\in[0,1]$ such that $\{x\}\times[0,1]\subseteq\bigcap\mathscr{B}_0$, so $\bigcap\mathscr{B}_0\nsubseteq U(x,C)$ for any non-empty $C\in\mathscr{C}_x$.


Your final attempt is first countable. Let $K=\left\{\left\langle\frac1n,\frac1n\right\rangle:n\in\Bbb Z^+\right\}\cup\{\langle 0,1\rangle\}$. For $k\in\Bbb Z^+$ let

$$U_k=\left(\{0\}\times\left(1-\frac1k,1\right]\right)\cup\left(\left(0,\frac1k\right)\times[0,1]\right)\;,$$

an open nbhd of $\langle 0,1\rangle$, and let

$$V_k=U_k\cup\bigcup_{n\ge k}\left(\left\{\frac1n\right\}\times\left(\frac1n-\frac1{kn},\frac1n+\frac1{kn}\right)\right)\;;$$

then $\{V_k:k\in\Bbb Z^+\}$ is an open nbhd base for the set $K$, so if $q$ is the quotient map, $\{q[V_k]:k\in\Bbb Z^+\}$ is an open nbhd base at the point corresponding to $K$ in $X/K$.

It’s the limit point $\langle 0,1\rangle$ that keeps this first countable and makes it behave differently in this respect from $\Bbb R/\Bbb N$. Unfortunately, you need to include it in $K$ to make the quotient Hausdorff. In order to make this idea work, you want to identify an infinite closed, discrete subset of the square to a point. $X$ doesn’t have such a subset, however. Suppose that $A=\{p_n:n\in\Bbb N\}$ is a countably infinite subset of $X$. For $n\in\Bbb N$ let $p_n=\langle x_n,y_n\rangle$. If there are an $x\in[0,1]$ and an infinite $M\subseteq\Bbb N$ such that $x_n=x$ for every $n\in M$, then clearly $A$ has an accumulation point in $\{x\}\times[0,1]$. If not, $\{y_n:n\in\Bbb N\}$ is infinite and therefore has an accumulation point $y$. But then at least one of the points $\langle y,0\rangle$ and $\langle y,1\rangle$ is an accumulation point of $A$.

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  • $\begingroup$ Sorry, confused my $T_0$ and $T_1$ axioms. Thanks for the clarification. I will need to mull over this for a bit, but thank you for tackling another problem of mine! Once I understand it I'll accept it. Also, just for additional clairfication, $\langle x,y\rangle$ is meant to denote the equivalence class containing $(x,y)$? $\endgroup$ – Hayden Nov 20 '13 at 22:58
  • $\begingroup$ @Hayden: You’re welcome. No, I use angle brackets for plain ordinary ordered pairs; it’s quite a common practice amongst set theorists, and I prefer it, since parentheses are awfully overloaded. $\endgroup$ – Brian M. Scott Nov 20 '13 at 23:00
  • $\begingroup$ Ah, I'll have to keep that in mind. Parentheses are quite common... Thanks for the clarification! Munkres uses $x\times y$ in lecture to make sure we don't confused ordered pairs and open intervals. $\endgroup$ – Hayden Nov 20 '13 at 23:01
  • $\begingroup$ @Hayden: My pleasure! Just leave a comment if you get stuck on anything. $\endgroup$ – Brian M. Scott Nov 20 '13 at 23:01

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