Is the Quotient Topology of Real Numbers Homeomorphic to Real Numbers?

[Korybut]

TL;DR Summary

R without interval and open interval

Hello!

I have two related exercises I need help with

1. Partition the space $\mathbb{R}$ into the interval $[a,b]$, and singletons disjoint from this interval. The associated equivalence $\sim$ is defined by $x\sim y$ if and only if either$x=y$ or $x,y\in[a,b]$. Then $\mathbb{R}/\sim$ is the space obtained from $\mathbb{R}$ by shrinking $[a,b]$ to a point. The space $\mathbb{R}/\sim$ looks like $\mathbb{R}$ show that it is homeomorphic to $\mathbb{R}$.

2. Suppose we use the open interval $(a,b)$ in place of $[a,b]$ in the previous excercise. So, in this case $x\sim y$ if and only if either $x=y$ or $x,y\in (a,b)$. Show that $\mathbb{R}/\sim$ is not homeomorphic to $\mathbb{R}$.

Concerning the first excercise I have the following
Canonical projection maps initial $\mathbb{R}$ to $\{ (-\infty,a),point,(b,+\infty)\}$. Here $point$ is the image of the interval $[a,b]$. It is not had to provide a bijective map with the following properties
$$ f:(-\infty,a)\rightarrow (-\infty,0),\;\; f:point\rightarrow 0,\;\; f:(b,+\infty)\rightarrow (0,+\infty)$$
How to show that this map is continuous based on definition of open sets in quotient topology?

Concerning the second excercise
How to show that something is not homeomorphic? Maybe there is some sort of criterion, cause to show that one can not invent homeomorphism looks like a tough problem.

 

[Infrared]

The singleton in part 2 corresponding to $(a,b)$ is open, which is not true of any point in $\mathbb{R}.$

For part 1 (and in general for identifying quotients), I find like to argue from the universal property of quotients: given $X\to X/\sim$ and a map $g:X\to Y$ such that $g(x_1)=g(x_2)$ whenever $x_1\sim x_2$, there is a unique map $(X/\sim)\to Y$ making the appropriate diagram commute, and this characterizes quotients $X\to X/\sim.$

Then you could argue that a continuous map $\mathbb{R}\to\mathbb{R}$ that is say, linear of slope $1$ away from $[a,b]$ and constant on $[a,b]$ satisfies the universal property above, so $(\mathbb{R}/\sim)\cong\mathbb{R}.$

Your method also works and is more direct. You would take open intervals in $\mathbb{R}$ and look at their preimages in $\mathbb{R}/\sim$ and determine that they're open according to the quotient topology (which means looking at their preimages again, in $\mathbb{R}$), but then you would also have to do the same for the inverse of your map to show that it is a homeomorphism.

 

[fresh_42]

Summary:: R without interval and open interval

Hello!

I have two related exercises I need help with

1. Partition the space $\mathbb{R}$ into the interval $[a,b]$, and singletons disjoint from this interval. The associated equivalence $\sim$ is defined by $x\sim y$ if and only if either$x=y$ or $x,y\in[a,b]$. Then $\mathbb{R}/\sim$ is the space obtained from $\mathbb{R}$ by shrinking $[a,b]$ to a point. The space $\mathbb{R}/\sim$ looks like $\mathbb{R}$ show that it is homeomorphic to $\mathbb{R}$.

2. Suppose we use the open interval $(a,b)$ in place of $[a,b]$ in the previous excercise. So, in this case $x\sim y$ if and only if either $x=y$ or $x,y\in (a,b)$. Show that $\mathbb{R}/\sim$ is not homeomorphic to $\mathbb{R}$.

Concerning the first excercise I have the following
Canonical projection maps initial $\mathbb{R}$ to $\{ (-\infty,a),point,(b,+\infty)\}$. Here $point$ is the image of the interval $[a,b]$. It is not had to provide a bijective map with the following properties
$$ f:(-\infty,a)\rightarrow (-\infty,0),\;\; f:point\rightarrow 0,\;\; f:(b,+\infty)\rightarrow ( 0,+\infty)$$
How to show that this map is continuous based on definition of open sets in quotient topology?

We have to show that $f$ and $f^{-1}$ are continuous. The open sets in $\mathbb{R}$ are unions of open intervals, the empty set and $\mathbb{R}$ itself. What are the open sets in $\mathbb{R}/\sim$? They are all sets $\{V\subseteq \mathbb{R}/\sim\,|\,\pi^{-1}(V)\subseteq \mathbb{R} \text{ open }\}$ where $\pi\, : \,\mathbb{R}\longrightarrow \mathbb{R}/\sim$ is the projection.

In order for $f$ to be continuous, we have to show that $f^{-1}(U)=\{y\in \mathbb{R}/\sim\,|\,f(y)\in U\}$ is open in $\mathbb{R}/\sim$ whenever $U\subseteq \mathbb{R}$ is open, i.e. that $\pi^{-1}(f^{-1}(U))\subseteq \mathbb{R}$ is open. We already know that $f$ is bijective, so we know that $\operatorname{im}(f)=\mathbb{R}$ and we can assume $U\subseteq \operatorname{im}(f).$

Since $f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V)$ we only need to consider open intervals, not their unions. So let's say $U=(c,d)\subseteq \mathbb{R}.$

... let me sort this out ... I'm typing directly without script ... you used "point" for a specific point in $\mathbb{R}/\sim$ which was a bit unlucky ...
\begin{align*}
\mathbb{R}\stackrel{\pi}{\longrightarrow } \mathbb{R}/\sim &\stackrel{f}{\longrightarrow } \mathbb{R} \stackrel{\pi}{\longrightarrow } \mathbb{R}/\sim \\
\{point\}& \longmapsto 0\\
f^{-1}(U)&\longrightarrow U \text{ open}
\end{align*}
We use the fact that $(c,d)=U\subseteq \mathbb{R}=\operatorname{im}(f)$. The intervals in the image of $f$ are identical intervals if they lie completely left or right of $\{0\}$. If $0\in U$ then $f^{-1}(U)=(c',d')$ for some $c',d'\in \mathbb{R}$ with $c'<a$ and $d'>b.$ Equality is excluded since $U$ is an open interval. So in any case $f^{-1}(U)$ has the form $(c',d')\subseteq \mathbb{R}/\sim.$ This means that $\pi^{-1}(f^{-1}(U))=\pi^{-1}((c',d'))=(c',d')\subseteq \mathbb{R}$ is open in $\mathbb{R}$ so $f^{-1}(U)$ is open in $\mathbb{R}/\sim$ by the definition of the quotient topology.

Note that we did not say that $(c',d') \subseteq \mathbb{R}/\sim$ are open because the boundaries are not included. We concluded that they are open because their pre-images under $\pi$ are open intervals in $\mathbb{R}.$ This is an essential difference. We showed they are open because they belong into the topology, the set of open sets, not because they "look" open.


Now you have to do the same for the function $f^{-1}.$ For it's pre-images we have $(f^{-1}(U))^{-1}=U.$

Concerning the second excercise
How to show that something is not homeomorphic? Maybe there is some sort of criterion, cause to show that one can not invent homeomorphism looks like a tough problem.

A set is open if and only if every point has an open neighborhood that is entirely contained in the set. (You can prove this by using that arbitrary unions of open sets are open.)

In case we consider $\mathbb{R}/(a,b)$ I suggest to investigate the point $\bar{x}:=\{a\}.$ Every point in $\mathbb{R}$ has an open neighborhood entirely contained in $\mathbb{R},$ of course. But does $\bar{x}:=\{a\}$ have?

 

[pasmith]

Summary:: R without interval and open interval

Hello!

I have two related exercises I need help with

1. Partition the space $\mathbb{R}$ into the interval $[a,b]$, and singletons disjoint from this interval. The associated equivalence $\sim$ is defined by $x\sim y$ if and only if either$x=y$ or $x,y\in[a,b]$. Then $\mathbb{R}/\sim$ is the space obtained from $\mathbb{R}$ by shrinking $[a,b]$ to a point. The space $\mathbb{R}/\sim$ looks like $\mathbb{R}$ show that it is homeomorphic to $\mathbb{R}$.

2. Suppose we use the open interval $(a,b)$ in place of $[a,b]$ in the previous excercise. So, in this case $x\sim y$ if and only if either $x=y$ or $x,y\in (a,b)$. Show that $\mathbb{R}/\sim$ is not homeomorphic to $\mathbb{R}$.

Concerning the second excercise
How to show that something is not homeomorphic? Maybe there is some sort of criterion, cause to show that one can not invent homeomorphism looks like a tough problem.

A homeomorphism is necessarily a continuous injection.

A continuous injection $f: \mathbb{R}/{\sim} \to \mathbb{R}$ must map $\{ \{x\}: x \leq a\}$ and $\{ \{x\}: x \geq b\}$ to disjoint unbounded closed intervals. That leaves us with an open interval of positive width into which we must place the single point $f((a,b))$ and then we're left with two open intervals of positive width which are not in the image of $f$. Since $f$ is not a surjection it cannot be a homeomorphism.

 

[Korybut]

A set is open if and only if every point has an open neighborhood that is entirely contained in the set. (You can prove this by using that arbitrary unions of open sets are open.)

In case we consider $\mathbb{R}/(a,b)$ I suggest to investigate the point $\bar{x}:=\{a\}.$ Every point in $\mathbb{R}$ has an open neighborhood entirely contained in $\mathbb{R},$ of course. But does $\bar{x}:=\{a\}$ have?

I believe I do not understand your exercise. I need to provide open set $U_a\subset\mathbb{R}/\sim$ which contains equivalence class of $a$. I can take $U_a=\pi((a-\varepsilon,a+\varepsilon))$, $\pi^{-1}(U_a)=(a-\varepsilon,b)$. Then $U_a$ is open due to properties of canonical projection. Or I misunderstand your exercise completely?

A homeomorphism is necessarily a continuous injection.

A continuous injection $f: \mathbb{R}/{\sim} \to \mathbb{R}$ must map $\{ \{x\}: x \leq a\}$ and $\{ \{x\}: x \geq b\}$ to disjoint unbounded closed intervals. That leaves us with an open interval of positive width into which we must place the single point $f((a,b))$ and then we're left with two open intervals of positive width which are not in the image of $f$. Since $f$ is not a surjection it cannot be a homeomorphism.

I do understand how contradiction emerges in this example. But I think very specific class of possible $f$s was considered, am I right?

p.s. Please forgive me such a delayed reply.

 

[Korybut]

I think I found rigorous way to solve second excercise. Homeomorphism should map closed sets to closed sets. Every single point in $\mathbb{R}$ is closed while $\pi((a,b))$ is open in $\mathbb{R}/\sim$ so there is no homeomorphism.

 

[pasmith]

I do understand how contradiction emerges in this example. But I think very specific class of possible $f$s was considered, am I right?

p.s. Please forgive me such a delayed reply.

There are more general possible continuous injections, but they have no hope of being surjections. We want $f$ to fail to be surjective for a specific reason, not because e.g. $f(\mathbb{R}/{\sim})$ is bounded.

But there is a better proof: If $f^{-1}$ is continuous then $f(U)$ must be open for every open $U \subset \mathbb{R}/{\sim}$. But what happens when $p^{-1}(U) = (a,b)$?