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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?


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.

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"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 19 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.

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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

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