Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is a consequence of Baire's theorem that a connected, locally connected complete space cannot be written $$ X = \bigcup_{n \geq 1}\ F_n$$ where the $F_n$ are nonempty, pairwise disjoint closed sets.

Does anyone know of a counter-example to this if we don't assume the space to be locally connected?

share|improve this question
    
How many sets $F_n$ do you want? Countably infinitely many? –  Rasmus Jan 5 '12 at 16:01
    
Just a trivial remark to avoid further erroneous answers: it must be an infinite disjoint union. A finite union won't work because of connectedness. –  t.b. Jan 5 '12 at 16:05
1  
To be clear: a complete connected non-locally connected metric space that can be written as a countable union of pairwise disjoint closed sets is sought, right? –  t.b. Jan 5 '12 at 16:17
    
@t.b. : that's right. Also one more remark: the space can't be compact, because of a theorem of Sierpinski. So if we look for some part of euclidean space it shouldn't be bounded. –  timofei Jan 5 '12 at 19:47
1  
You use the word complete, but you don't say metric space. Do you mean a complete metric space? –  JDH Jan 6 '12 at 0:27

1 Answer 1

up vote 5 down vote accepted

If $e_n$ are the standard unit vectors in $\ell^2$, let $F_j$ consist of line segments from $e_j$ to $(1/j) e_j + e_k$ for $1 \le k < j$ ($F_1$ is the single point $e_1$). Then the $F_j$ are disjoint, closed and connected, and their union is closed and connected.

share|improve this answer
    
Indeed! Thank you very much. –  timofei Jan 6 '12 at 14:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.