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quasar987

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## Homework Statement

Our professor gave us a few exercises at the end of class the other day and it is possible that he wrote the problem wrong, or I copied it wrong, but in any case, something's fishy about it. It says

*Show that there exists a topology on the set [itex]\mathcal{D}[/itex] of all lines of [itex]\mathbb{R}^2[/itex] that makes it homeomorphic to the circle [itex]\mathbb{S}^1[/itex].*

## The Attempt at a Solution

I think I found a proof that no such topology can exists on the simple ground that no bijection can exist btw [itex]\mathcal{D}[/itex] and [itex]\mathbb{S}^1[/itex]. It goes like this...

1° Note that [itex]\approx[/itex] denoting isomorphisms of sets (i.e. [itex]A\approx B[/itex] means there exists a bijection btw A and B) is an equivalence relation on the set of all sets.

2° [itex]\mathbb{S}^1[/itex] as a set is isomorphic to [itex][1,0)[/itex] via, say, [itex]t\mapsto e^{2\pi it}[/itex].

3° [itex]\mathcal{D}\approx \mathbb{R}^2[/itex] via [itex]\{(x,mx+b): x\in\mathbb{R}\}\mapsto (m,b)[/itex]

4° [itex][[1,0)]_{\approx}\neq [\mathbb{R}^2]_{\approx}[/itex]. So because [itex]\approx[/itex] is an equivalence relation, [itex][\mathcal{D}]_{\approx} \neq [\mathbb{S}^1]_{\approx}[/itex], i.e. there can exists no bijection btw these two sets, let alone homeomorphisms.

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