Is the Intersection of Compact Connected Sets Always Connected in Metric Spaces?

[quasar987]

Homework Statement

Show that if {K_n} is a decreasing family of compact connected sets in a metric space, then their intersection is connected as well. Illustrate with an example why 'compact' is necessary instead of just 'closed'.

The Attempt at a Solution

Well, I have a example for the second part of the question. Consider F_n = R²\{(x,y): -n<y<n, -1<x<1}. Then each F_n is closed and (path-)connected, but their intersection is the plane separated in half along the y-axis by this open band of width 2, which is not connected.

For the first part though, I can visualize why it's true for simple examples, but I don't know how to approach a general proof.

 

[Dick]

Ok. Suppose the limit set K is disconnected. That means there are two open sets A and B that disconnect K, right? So A intersect B is empty but K is contained in AUB. Since K is compact we can define A and B so that A, B and K are all contained in the interior of a closed ball R. Consider the compact set R-(AUB). K_n is connected so it must intersect R-(AUB). So for each K_n there is a point in K_n, say x_n, contained in the compact set R-(AUB). I'll let you take over now...

 

[quasar987]

Nice, thx.