# Can a nontrivial quotient space of R be homeomorphic to R?

1. Mar 19, 2013

### lugita15

Intuitively, one would assume that the quotient space of a topological space under an equivalence relation would always be smaller than the original space. It turns out this is not remotely true. I'm specifically interested in quotient spaces of ℝ (under the standard topology).

We can easily make a quotient space of ℝ be homeomorphic to ℝ, for instance be gluing all the points in an interval into a single point. We can even glue infinitely many intervals into points, and still get a quotient space homeomorphic to ℝ. But my question is this: let us call an equivalence relation "nontrivial" if every equivalence class has at least two elements. Then does there exist a nontrivial equivalence relation on ℝ such that the quotient space is homeomorphic to ℝ?

Any help would be greatly appreciated.

2. Mar 20, 2013

### micromass

Staff Emeritus
There is a theorem that if $f:X\rightarrow Y$ is a closed (or open) continuous surjection, then it is a quotient map. So if we define for $x,y\in X$ the following equivalence relation

$$x\sim y~\Leftrightarrow ~ f(x)=f(y)$$

then we have that $Y=X/\sim$.

So if we succeed to find a closed (or open) continuous surjection $f:\mathbb{R}\rightarrow \mathbb{R}$, then we will have found an equivalence relation such that its quptient is $\mathbb{R}$.

I think that the following

$$f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x\sin(x)$$

is a closed and continuous surjection. So this would be an example.

3. Apr 2, 2013

### lugita15

Thanks micromass. What if we imposed an additional condition on the equivalence relation: if a<b<c and a~c, then a~b. Under that condition, it's impossible to have the quotient space be homeomorphic to ℝ, right?

4. Apr 4, 2013

### micromass

Staff Emeritus
Yeah, I think it should impossible then. But let me think of a proof...

5. Apr 5, 2013

### Amir Livne

With the additional constraint, the set of image points with more than one origin cannot be uncountable, since each preimage contains an open interval. From separability of $\mathbb{R}$ there is no uncountable set of pairwise disjoint open intervals.