Relation between subspace topology and quotient topology on a space.

general-topology

Let $S$ be the set $\mathbb{R}^2\setminus \{(0, y)\ |\ y\neq 0\}$ obtained by removng the $y$-axis except for the origin. Let $\tau_1$ denote the subspace topology on $S$ induced from the usual topology of $\mathbb{R}^2$. Next, consider the surjective map $\pi:\mathbb{R}^2\to S$ defined by $$(x, y)\mapsto (x, y)\ \text{for}\ x\neq 0, \ \text{and}\ (0, y)\mapsto (0, 0)\ \text{for all}\ y.$$
Let $\tau_2$ be the resulting quotient topology on $S$. Then show that

  1. $\tau_1$ and $\tau_2$ are comparable.

  2. $\tau_2$ is regular.

  3. $\tau_2$ is meterizable.

For the first option, I need to show that whether $\tau_1\subset \tau_2$ or vice-versa.
By definition, $\tau_1$ is the finest topology such that $i:S\to \mathbb{R}^2$ (inclusion) is continuous and $\tau_2$ is the finest topology for which $\pi$ is continuous. I do not understand how to compare them. Can anyone please explain it.

For option 2, I need to show that corresponding to a closed set $C$ in $\tau_2$ and a point not in $C$, there exist disjoint open sets that contains them. After taking a closed set $C$ and $p\notin C$, I get that $p\in C^c$, where $C^c$ is an open set. This means there exists an open set $O_p\in \tau_2$ such that $$p\in O_p\subset C^c.$$ How to move ahead.

For option 3, I think of applying Uryshon lemma. Is it right approach?

Best Answer

(Getting another question off the unanswered list; I will as usual leave a little for the reader to do.)

For this problem the characterizations of the subspace and quotient topologies given in the question are not especially useful; the ones suggested by J. Damé in the comments are much more helpful, because they give explicit descriptions of the open sets in these topologies.

Let $\tau$ be the usual topology on $\Bbb R^2$, and for convenience let $p=\langle 0,0\rangle$. Let $U\subseteq S$, and suppose first that $p\notin U$.

  • Show that $U\in\tau_1$ iff $U\in\tau$ iff $U\in\tau_2$.

Now suppose that $p\in U$.

  • Show that $U\in\tau_1$ iff there is a $V\in\tau$ such that $p\in V$ and $V\cap S=U$.
  • Show that $U\in\tau_2$ iff there is a $V\in\tau$ such that $\{0\}\times\Bbb R\subseteq V$ and $V\cap S=U$. Show further that $\operatorname{cl}_{\tau_2}U=S\cap\operatorname{cl}_\tau V$; this will be useful in showing that $\tau_2$ is regular.

These observations let us determine exactly what sets are in $\tau_1$ and what sets are in $\tau_2$. In particular it should be clear that $\tau_2\subsetneqq\tau_1$, since every $\tau$-open nbhd of the $y$-axis is a $\tau$-open nbhd of the origin, but there are clearly $\tau$-open nbhds of the origin that do not contain the $y$-axis.

Every metrizable space is regular, so we’d be done if we could show that $\tau_2$ is metrizable. Unfortunately, it is not: it is not even first countable. Specifically, the space $\langle S,\tau_2\rangle$ does not have a countable local base at the point $p$.

To see this, suppose that $\mathscr{B}=\{B_n:n\in\Bbb Z^+\}$ is a family of open nbhds of $p$. For each $n\in\Bbb Z^+$ there is a $k_n\in\Bbb Z^+$ such that $\left\langle\frac1{k_n},n\right\rangle\in B_n$. Let $F=\left\{\left\langle\frac1{k_n},n\right\rangle:n\in\Bbb Z^+\right\}$, $V=\Bbb R^2\setminus F$, and $U=V\cap S$. $F$ is closed in $\left\langle\Bbb R^2,\tau\right\rangle$ and disjoint from the $y$-axis, so $V$ is a $\tau$-open nbhd of the $y$-axis, and $p\in U\in\tau_2$. But $\left\langle\frac1{k_n},n\right\rangle\in B_n\setminus U$ for each $n\in\Bbb Z^+$, so $B_n\nsubseteq U$ for each $n\in\Bbb Z^+$, and $\mathscr{B}$ is therefore not a local base at $p$.

The topology $\tau_1$, on the other hand, is metrizable: it’s easy to check that if $d$ is any metric on $\Bbb R^2$ that generates the usual topology $\tau$, then $d\upharpoonright(S\times S)$ is a metric on $S$ that generates the subspace topology $\tau_1$.

Finally, $\tau_2$ is regular. It suffices to show that if $x\in S$, and $x\in U\in\tau_2$, then there is a $G\in\tau_2$ such that $x\in G\subseteq\operatorname{cl}_{\tau_2}G\subseteq U$. This is easy to see if $x\ne p$, so suppose that $p\in U\in\tau_2$. There is a $V\in\tau$ such that $\{0\}\times\Bbb R\subseteq V$ and $V\cap S=U$. $\left\langle\Bbb R^2,\tau\right\rangle$ is normal, and $\{0\}\times\Bbb R$ is $\tau$-closed, so there is a $W\in\tau$ such that $\{0\}\times\Bbb R\subseteq W\subseteq\operatorname{cl}_\tau W\subseteq V$. Let $G=W\cap S$; then $G\in\tau_2$, and

$$p\in G\subseteq\operatorname{cl}_{\tau_2}G=S\cap\operatorname{cl}_\tau W\subseteq S\cap V=U\,,$$

as desired.

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