What is the connectedness problem from Conway's Complex Analysis?

[rudinreader]

From Conway's Complex Analysis, page 17, 2.2.5:

Suppose $$F \subseteq X$$ is closed and connected. If a,b are in F and e > 0, then there exists $$a = z_0,z_1,...,z_n = b$$ such that $$d(z_{k-1},z_k) < e$$ for k in {1,...,n}.

I don't see the answer to this off the top of my head.. anyone else see it? Is this a special case of a more general idea?

 

[Ben Niehoff]

It means that in a connected, closed region, you can get from point A to point B by taking arbitrarily small steps, within the region. In fact, it seems like you could even take this as the definition of "connected", as far as I can tell. It seems to be saying, "There is a continuous path in F from A to B".

 

[rudinreader]

But satisfying the property above doesn't imply connectedness. Here are some examples(and the last one is closed):
Q, the rationals
$$[0,1) \cup (1,2]$$
$$\{(x,\frac{1}{|x|} ): x < 0\} \cup \{(x,\frac{1}{|x|} ): x > 0 \}$$

 

[Ben Niehoff]

True...the sequence being described isn't actually continuous. The connectedness property does imply, however, the existence of a sequence of arbitrarily-close points which begins at A and ends at B.

I thought this was interesting and decided to read a bit about connectedness, as topology isn't really my specialty. Apparently there is a distinction between connectedness and path-connectedness. Path-connectedness is what I described, which is not quite what your text describes, so I was wrong. There also exist, it seems, certain pathological cases which are connected but not path-connected.

The "sequence-connected" version you have should be true of all connected spaces, though. However, the converse is not true. I'm not sure what you're asking...were you trying to prove it?

 

[rudinreader]

OK I got it.

Ok here's a proof. It's not as hard as I thought. Fix e > 0, and a,b in X.
Let $$A = \{z \in X: there \ exists \ a = z_0, ..., z_n = z \ in \ X \ such \ that \ d(z_{k-1},z_k) < e \}$$.
We have a in A, so A is not empty. To prove b in A, it suffices to prove A is open and closed in X, hence A = X.
To do that, you suppose $$z \in A$$. Then choose a sequence $$a = z_0,..., z_n = z$$ such that $$d(z_{k-1},z_k) < e$$. Then for any y in $$B(e,z)$$, just set y = $$z_{n+1}$$. That shows y in A, hence A is open in X.

To show A is closed, you choose a point z in X\A, and using a similar argument (of appending to the end a sequence), show that there is no sequence connecting a to any point of a neighborhood of z.

 

[rudinreader]

Yes, I was trying to prove it.. and in the proof I just threw away the F, and considered a connected metric space X. So that's more general. A connected closed subset F is of course a connected metric space.

 

[Ben Niehoff]

I'm not sure what your proof is proving. We already know that a and b are in F; we're trying to prove that a sequence exists between a and b whose elements z_i are arbitrarily close together.

It seems to me that this should follow from the fact that F is connected. I don't see how F being closed figures into it at all...(and in fact, as your earlier examples showed, F doesn't even have to be connected for the sequence to exist).

 

[rudinreader]

What I did was fix e > 0, and defined a set A to be the z in X that has such an "e-path" to a. I showed b was in A (because A=X). Yes, I threw out the closed part. In the proof, I don't mention F, I just look at a connected metric space X.

 

[HallsofIvy]

True...the sequence being described isn't actually continuous. The connectedness property does imply, however, the existence of a sequence of arbitrarily-close points which begins at A and ends at B.

I thought this was interesting and decided to read a bit about connectedness, as topology isn't really my specialty. Apparently there is a distinction between connectedness and path-connectedness. Path-connectedness is what I described, which is not quite what your text describes, so I was wrong. There also exist, it seems, certain pathological cases which are connected but not path-connected.

The "sequence-connected" version you have should be true of all connected spaces, though. However, the converse is not true. I'm not sure what you're asking...were you trying to prove it?

Any path-connected set is connected, any open connected set is path connected, any closed connected set is path connected.

Here's a puzzle I concocted several years ago: Find 2 sets P and Q, both contained in the closed square in R² with vertices at (1,1), (-1,1), (-1,-1), and (1, -1) such that:
a) P contains (1,1) and (-1,-1) while Q contains (-1, 1) and (1, -1)
b) P and Q are both connected sets
c) P and Q are disjoint.

This can be done by defining P and Q as the graphs of two functions such that the graphs are connected but NOT path connected.

 

[rudinreader]

any closed connected set is path connected.

I assume that's a typo. Because the "closed topologists sine curve" is a closed connected set that is not path connected.

Your puzzle is interesting, but I don't see the trick in constructing the said P and Q. Any more hints without giving it away?

 

[HallsofIvy]

I wish it were a typo! I just plain mis-spoke. Thanks for the correction.

Make the middle part of each path a variation on y= sin(1/x) (is that the "topologists sine curve" you were referring to?). multiply by, say, 0.95 to keep it inside the square and add or subtract 0.03 to get them "miss" each other.