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.

Given a descending sequence of sets

$$ F_1\supset F_2\supset\cdots F_n\supset\cdots $$ in which each $F_i$ is connected. I wonder if the limit set

$$ F=\bigcap_{i=1}^\infty F_i $$ is still connected? I believe it is, but cannot make a proof. Anyone can help?


Updated:

Samuel has showed a counter example.

Thus now I wonder can I add some constraints such that the conclusion holds?

I ask this problem because when I look up the The Princeton Companion to Mathematics,chapter IV.14. Dynamics, section 2.8 The Mandelbrot Set M, there is the following words:

It follows from the above that as $t$ approaches zero, the equipotential of potential $t$, together with its interior, gets closer and closer to M: that is, M is the intersection of all such sets. Hence, M is a connected, closed, bounded subset of the plane.

I wonder why such argument shows $M$, the Mandelbrot set, is connected.

share|improve this question
    
"Limit" seems like a misleading word. "Intersection" would be better. A point is in the intersection iff it's in all of the sets. That's a simpler definition than the usual definitions of limits. –  Michael Hardy Jun 21 '13 at 0:07
    
@Michael Hardy : If you see a space $X$ with a partial order defined by inclusion, this is indeed a limit in the partial order sense : $$ \liminf_{n \to \infty} A_n \overset{def}= \bigcup_{n \in \mathbb N} \bigcap_{k \ge n} A_k, \qquad \limsup_{n \to \infty} A_n \overset{def}= \bigcap_{n \in \mathbb N} \bigcup_{k \ge n} A_k $$ and when $\liminf = \limsup$, we call it the limit of the sequence $A_n$. In the case of a monotone sequence the limit always exists. So it's not really misleading. –  Patrick Da Silva Jun 21 '13 at 0:46
    
@PatrickDaSilva : But here you're relying on the order relation on $\mathbb N$, whereas the set $\bigcap_{k\in S} A_k$ is defined without any order or other structure on the set $S$, simply by saying $x\in\text{this set}$ iff $\forall k\in S,\ x\in A_k$. That's simpler than any definition of a limit. –  Michael Hardy Jun 21 '13 at 17:12

1 Answer 1

up vote 18 down vote accepted

No. Let $F_n$ be the the plane $\mathbb R^2$ minus the line $\{0\}\times(-\infty,n)$.

Added: It is true when all the $F_n$ are compact subsets of $\mathbb R^N$. Suppose otherwise: then there exist open disjoint sets $A,B$ such that $F$ contains points of both $A$ and $B$ and $F$ is contained in $A\cup B$. Now consider $F_n\cap (\partial A)$. Since each $F_n$ is connected, and contains points in both $A$ and $B$, the intersection $F_n\cap (\partial A)$ must be nonempty, and moreover, for $n=1,2,3,\ldots$ it is a decreasing sequence of compact sets, and therefore the intersection of all $F_n\cap (\partial A)$ is nonempty. Contradiction. Thus $F$ is connected.

share|improve this answer
    
good counter example, thank you! Furthermore, I wonder can I add some constraints such that the conclusion holds? For example, what about each $F_i$ is furthermore compact? –  hxhxhx88 Jun 14 '13 at 16:55
    
@hxhxhx88: Yes, it is true when all $F_n$ are compact; see my edit. –  Samuel Jun 14 '13 at 17:09
    
Thank you soooo much! BTW, when you arguing the intersection of all $F_n\cap(\partial A)$ is nonempty, do you use the Cantor's intersection theorem? If so, such conclusion is only valid in $\mathbb{R}^n$. Or is there a general consequence that limit of decreasing non-empty compact set is non-empty? –  hxhxhx88 Jun 15 '13 at 4:45
    
@hxhxhx88: You are right. Cantor's intersection theorem can fail: if we choose the sets $(0,1)\supseteq (0,1/2)\supseteq (0,1/3)\supseteq \cdots$, they are compact with respect to the trivial topology, yet have empty intersection. –  Samuel Jun 15 '13 at 10:45
    
@Samuel : Something came up to my mind... if they're closed, then they won't be connected in the topogical space $X$, but they will be in its one-point compactification, right? The subsets $F_n$ will remain closed in the compactification and thus become compact there, where your argument holds. In the case of $\mathbb R^2$, it's the "point at infinity" which preserves the connectedness. Apparently this kind of Hausdorff compactification (which I need for 'closed implies compact' statement) works if and only if the topological space is Tychonoff. –  Patrick Da Silva Jun 21 '13 at 1:00

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.