Cute Proofs: Closure of a Connected Subspace is Connected.
May 16, 2011
[Big Edit: Thank you to the commenters below (and especially the latest one) who pointed out some of my previous statements only held true in a specific case. I’ve streamlined the proof, and I think everything is kosher now. I’ve also added some counterexamples to the end.]
I was reading through some of the earlier chapters of Bredon and I got to one of those questions that should be easy but I couldn’t recall the argument off of the top of my head. The actual question is slightly more general, but the question we’ll answer first is:
Problem: If is connected, show is connected.
There are several nice definitions of connected to choose from. In particular, we have:
A subset is connected if one (and hence all) of the following hold:
- There DO NOT exist disjoint non-empty closed (with respect to ) sets such that and but .
- The only subsets of which are both open and closed in the relative topology are and .
[Note: In a former draft of this post, I had (incorrectly!) stated that a subset is connected if we did not have two disjoint open sets which do not intersect trivially, but which union to . This can be shown to be false (which was done by a clever commenter below), but it will work if (and therefore in the subspace topology) is a metric space.]
To solve this, we will use the first statement above but, for fun, try to prove it yourself using the second statement. We’ll be using the contrapositive, meaning we will prove if is not connected, then is not connected. This is equivalent to proving the theorem above (why?).
Solution. Let it be the case that is not connected, so that there exists disjoint closed sets neither of which intersect trivially. Suppose that is connected; then for any two disjoint closed sets with we have that one of these closed sets intersects trivially (or else it would not be connected). In other words, wlog we have . Taking the closure of both sides, we have , but since and are disjoint this implies that , contradicting the fact that neither nor may intersect nontrivially.
In fact, nearly the same kind of argument shows that if there is some such that with connected, then is also connected. This is pretty nice; it tells us we can add limit points to our connected space (even if we only add some limit points and not others!) and it remains connected.
Some counter examples…
One might ask, "Hey, what if is connected? Then is connected?" And a minute of thinking should provide a counter-example. If not, then just think about and its closure.
Additionally, one might think that if a set is connected, then its interior (if it is nonempty) would be connected. This is true in the case of the closed disk , but if we were to put two disjoint disks in the plane and connect them together with a line (this will look like a barbell, or a pair of eyeglasses) the interiors will be two disjoint open disks, which are clearly not connected.
Maybe, though, the boundary of a connected set is always connected (if it is nonempty). To do away with this idea, just think about the interval in .
A question to think about (assume all of these sets are non-empty):
- If the boundary of some set is connected, is connected? (For this one, think about a figure-8 in the plane.)
- If the interior of some set is connected, is connected? (For this one, think about what the interior of a point is, and then let be some nice set with non-empty interior union some point.)
- If the interior and boundary of some set are connected, is ? (For this one, you might have to think a bit. I’ll try to hide an answer a little bit so you can work on it.
Consider the set of positive rational numbers, union the interval . The interior will be which is connected, and the boundary will be which is connected, but the set itself is not connected (for the same reason the rationals are not connected).