# Classification of connected subsets of the real line (up to homeomorphism)

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?

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Yes. ${}{}{}{}{}$ –  Mariano Suárez-Alvarez Aug 7 '11 at 23:01
I assume you're asking if the given 3 intervals exhaust the classification. Seems so. Also, if you mean the diamond character, that means Mariano is a moderator. If you mean the split-second $\{\}$'s in the comment, that's just invisible markup to circumvent the character limit. –  anon Aug 7 '11 at 23:03
Singletons? {}{}{} I can't figure out how to do the bracket trick, however. –  user83827 Aug 7 '11 at 23:06
@user10: comments have to consist of at least 15 characters, and the comment I meant to write, «Yes.» has only four: so I added a ${}{}{}{}{}$ which is a little piece of LaTeX which results in nothing, but still counts when the software is counting characters. –  Mariano Suárez-Alvarez Aug 7 '11 at 23:09
user10: Mathematics require some level of maturity (mathematical maturity, that is) to be able and assess your own proofs. While there is nothing actually wrong in asking for proofing your proofs, you need to develop a sense of critical thinking and eventually be able to find the nitpicks by your own. This question is a good example for that, a fine proof that you were insecure about. Have confidence in your arguments, even if sometimes you will mistake, these mistakes will develop your critical thinking even more than proofing by others would. Trust me, I know. :-) –  Asaf Karagila Aug 7 '11 at 23:12

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.

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Brian, $\{x\}=[x,x]$ while $\varnothing=(x,x)$. I do agree that this needs to be mentioned. –  Asaf Karagila Aug 8 '11 at 5:23
@Asaf: I'm not sure I understand your comment (I mean, I acknowledge the truth of the first sentence, but I don't see what it has to do with anything, and I don't know what the second sentence is referring to). That came across more rudely than I intended; I'm genuinely confused. .@Brian: Like Mariano, I remember being taught that the empty space is not connected. I think the explanation was very similar to the justification that the empty graph is not connected -- conveniently, I've forgotten the justification for that as well. –  user83827 Aug 8 '11 at 15:50
@ccc: This is because the OP asked whether his proof of three types is correct, while Brian pointed out five types. I mentioned that the additional two are the open/closed cases of $a=b$ in the intervals. I also added that I agree that they need to be mentioned explicitly. I hope that clears things up. :-) –  Asaf Karagila Aug 8 '11 at 16:57
@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! –  user83827 Aug 8 '11 at 17:06