# Connectedness problem

1. Nov 27, 2007

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?

Last edited: Nov 27, 2007
2. Nov 28, 2007

### 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".

3. Nov 28, 2007

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

4. Nov 28, 2007

### 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?

5. Nov 28, 2007

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.

Last edited: Nov 28, 2007
6. Nov 28, 2007

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.

7. Nov 28, 2007

### 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).

8. Nov 28, 2007

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.

9. Nov 28, 2007

### HallsofIvy

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.

10. Nov 28, 2007

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?

11. Nov 28, 2007

### 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.