Show that the quotient space is not Hausdorf

Question

This is problem 2.13.3 in the Gamelin and Greene topology book.

1. Theorem

Define $ X = [0,1] \times [0,1] $ with equivalence relation $$ (s_0, t_0) \sim (s_1, t_1) \iff t_0 = t_1 > 0 $$ (1) describe the quotient space X/~.
(2) show that it is not Hausdorff.

2. My answer to (1)

Let \begin{align} \pi &: X \rightarrow X/\sim \\ E &:= \{ [0,1] \times {t} : t \in (0,1] \} \\ F &:= \{ (s, 0) : s \in [0,1] \} \end{align}

Then $E$ and $F$ represent collections the equivalence classes of X/~. Moreover, $$ \tag{1} X/\sim ~ = E \sqcup F $$

Note that the collection $F$ has isolated points as equivalence classes, while $E$ does not.

Proof from textbook

See page 219 in the Gamelin and Greene topology book.

As a point set, X/~ can be viewed as the disjoint union of a closed interval I and a semi-open interval J. Points of J have the usual neighborhood base, but points s of I have a neighborhood base of sets that are unions of open subintervals of I containing s and open intervals at the open end of J. Thus neighborhoods of any two points of I contain in common a subinterval at the open end of J.

3. Questions

3.1 Confusion about open/closed intervals in X/~

I don't understand how to characterize the open intervals of X/~. By the quotient topology, I "understand" that $$ \pi^{-1}(U) \in \tau_X \iff U \in \tau_{X/ \sim} $$ I think that $F$ is closed in X/~ because

$$ [0,1] \times \{0\} = \pi^{-1}(F) $$ is closed in X. But that implies that E (the book labels this one J) is open, i.e. not a half-open interval as the proof suggests.

3.2 "Points of J have the usual neighborhood base"

J is semi-open. What is the "usual neighborhood base"? I assume it means neighborhoods of points in J will be contained in J (even though J isn't open?).

3.3 "Points s of I have a neighborhood base of sets that are unions of open subintervals of I containing s and open intervals at the open end of J".

Since I is a closed interval, it makes sense that certain neighborhoods of points in I won't be contained in I. If the open neighborhood falls outside of I, it's clear that it must intersect with J.

But the proof claims that neighborhoods of points in I always fall outside of J:

"Thus neighborhoods of any two points of I contain in common a subinterval at the open end of J".

That doesn't make sense to me -- surely for any point in I that is not an endpoint, there exists a neighborhood contained in I.

Answer 1

Let's take a look at the book's solution: at the end the suggest points that do not have disjoint neighborhoods in the quotient, namely any two points in $I=[0,1]\times\{0\}$ (or rather their classes): let's fill in the details:

let $x = [(0,0)]$ (the equivalence class of $(0,0)$, i.e. $\{(0,0)\}$ as points on the bottom are only equivalent to themselves and $y=[(\frac12,0)] = \{(\frac12,0)\}$. Suppose $U$ is an open neighborhood of $x$ in the quotient, so that $q^{-1}[U]$ is an open subset of $[0,1] \times [0,1]$ that contains $(0,0)$. As we are in the usual (product) topology there we know that for some open intervals $[0,s)$ and $[0,t)$ we have that $$[0,s) \times [0,t) \subseteq q^{-1}[U]\tag{1}$$

Similarly if $V$ is an open neighbourhood of $y$, $q^{-1}[V]$ is a usual open neighbourhood of $(\frac12,0)$ in $[0,1]\times [0,1]$ so for some $0 < s_1 < \frac12 < s_2 < 1; t_2 < 1$ we have

$$(s_1,s_2) \times [0,t_2) \subseteq q^{-1}[V]\tag{2}$$

Note that $(1)$ implies that $[0,1]\times \{u\} = [(0,u)]=[(\frac12,u)] \in U$ for any $u < t$ and $(2)$ implies the same thing for all $u < t_2$.

As there are many $u$ obeying $0 < u < \min(t,t_2)$, we have that $U$ and $V$ are not disjoint but both contain all classes with small enough second coordinate. So $J= [0,1] \times (0,1]$ and both neighborhoods contain a small interval at the open end of the second coordinate of $J$, namely $(0,\min(t,t_2))$ (i.e. their classes).

Answer 2

The solution given in your book is correct but not very clear. I suggest that you use sequences instead. Recall that in Hausdorff spaces, a sequence has a single limit. Let us build a sequence that has two limits.

First we work on the square $X$ and consider the two sequences $$u_n = (0, 1/n) \rightarrow (0,0), \quad v_n = (1,1/n) \rightarrow (1,0)$$ The two limits are different. Now let us project the two sequences on the quotient. They become equal but their limits stay different. Note that the quotient map is continuous so that $$\pi(u_n) \rightarrow \pi(0,0), \quad \pi(v_n) \rightarrow \pi(1,0)$$ and $\pi(u_n)=\pi(v_n)$ for all $n$, together with $\pi(0,0) \neq \pi(1,0)$.