Identify all interior points of the unit square and explain the Quotient space obtained

Question

I have been given this excercise. Define in the unit square the following relation: $[(a,b)] = \{(x,y): 0<x<1, 0<y<1\}$ if (a,b) is interior. Otherwise, $[(a,b)] = \{(a,b)\}$ That is, we identify all interior points in one only point, and each point in the boundary is only identified with itself.

Then, the excercise asks to what set of $\mathbb R^3$ is homeomorphic this quotient space. But I think that is not possible, because the interior point is an open point in the quotient space, and subsets of $\mathbb R^3$ have no open points. I see the space as $S^1$ with a "generic point". Am I right?

Answer 1

A subset of $\mathbb{R}^3$ can have an open set which is only a point. For example, if you take all the elements of the form $(n, 0, 0)$ in $\mathbb{R}^3$ with the subspace topology, it’s homeomorphic to the discrete topology on $\mathbb{N}$, because you can find around any element a open set which doesn’t contain any other element.

But a subspace of $\mathbb{R}^3$ is always hausdorff, and your space is not, because any open set which contains a point in the boundary, contains an element in the interior. Therefore, any open set which contains a point in the boundary must contain the interior point. So there’s no answer to the exercise.

Answer 2

{ (0,0,0), (x,y,0) : x$^2$ + y$^2$ = 1 }