3
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I want to describe, up to homeomorphism, all the proper connected subsets of $\mathbb{R}$. I know the theorem that $A \subset \mathbb{R}$ is connected if and only if $A$ is an interval.

So consider the following intervals:

$[a,b), [a,b], (a,b)$

Note $[a,b)$ is not homeomorphic to $[a,b]$ (compactness argument) and $[a,b)$ is not homeomorphic to $(a,b)$ (by connectedness). Similary $[a,b]$ is not homeomorphic to $[a,b)$ (remove a,b from [a,b]).

Now any ray $(a,\infty)$ is homeomorphic to $(1,\infty)$ which is homeomorphic to $(0,1)$ and $(0,1)$ is homeomorphic to $(a,b)$. Similarly the ray $[-a,\infty)$ is homeomorphic to $(0,1]$. So I think this covers all the cases.

So I believe the answer is $3$, yes?

  • 0
    @Mariano: OK, I'll remain agnostic for now, having no real reason to hold one position or the other. (The reason I deleted my comment is that an answer was posted making it redundant before your comment was posted. However, for some reason that answer was deleted.)2011-08-07

1 Answers 1

5

Not quite. It’s actually five:

  1. the type of $(0,1)$ (and all open intervals and rays, including $\mathbb{R}$ itself);
  2. the type of $[0,1)$ (and all half-open intervals and closed rays);
  3. the type of $[0,1]$ (and all non-trivial closed intervals);
  4. the type of $\{0\}$ (and all singletons); and
  5. the type whose only representative is the empty set.

Those like Mariano who exclude $\varnothing$ from the class of connected sets will have only the first four types.

  • 0
    @Asaf: Oh, ok. It looked to me like you were claiming something along the lines of $[0,1]$ is homeomorphic to $\{0\}$ because you can write $\{0\}$ as a closed interval. While that's something I could see myself doing, I had a feeling you meant something slightly more reasonable :-). Thanks for clearing up my confusion!2011-08-08