Connected and Path Connected

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March 14th 2010, 10:54 AM
Showcase_22

Connected and Path Connected

Quote:

Which of the following subsets of $\mathbb{R}^2$ are a). path connected b). connected:

i). $B_1((1,0)) \cup B_1((-1,0))$

ii). $\overline{B_1((1,0))} \cup \overline{B((-1,0))}$

iii). $\overline{B_1((1,0))} \cup B_1((-1,0))$

I think that:

i). I don't think this is path connected. If you took a point in each set , for example $\frac{1}{2}$ and $-\frac{1}{2}$ , you couldn't get from one to the other in a continuous path. This is because you would have to travel through 0, but 0 isn't in either set.

ii). I think this is path connected. Since both sets now contain 0, you could travel between them.
Since they are path connected, the subsets are connected.

iii). I think this is path connected. One set contains 0, the other contains points arbitrarily close to 0. Therefore you could make a continuous map between the two.
Once again, since they are path connected they are connected.

P.S: Thanks to everyone who has helped me with any topology questions over the past few days. You've helped me so much!
March 14th 2010, 11:52 AM
Tinyboss

Watch out! Your conclusions are correct, but your reasoning is insufficient.

For (1), think carefully about the definition of connectedness.

You (should) know that balls are connected, so (2) is two connected spaces with nonempty intersection, and (3) is two connected spaces each of which intersects a third connected space (think about the path you'd like to draw). There are theorems you can cite.

Edit: I wasn't as clear as I could have been on (3). The reason this one isn't as straightforward as (2) is that the two balls have empty intersection. But onsider the set $P=[-1,1]\times\{0\}$ , which is a straight-line path from the center of one ball to the center of the other. If you call the two balls $B_1,B_2$ , then you can write $B_1\cup B_2=B_1\cup B_2\cup P$ , and then apply the theorem about connected spaces with nonempty intersection.
March 14th 2010, 12:24 PM
Showcase_22

I quite like what you've said about 2) and 3). I went back to my book and saw the theorems about 2) that you mentioned. Your edit for 3) was great!

I'm not sure what you mean about 1) though. Do you mean like this:

Let $U=B_1((1,0))$ and $V=B_1((-1,0))$ .

$U,V$ are both open.

$T \subset U \cup V$ , $U \cap V= \phi$ , $U \cap T \neq \phi$ and $V \cap T \neq \phi$ .

According to my notes, since all these conditions are satisfied the set is disconnected.
March 14th 2010, 12:28 PM
Tinyboss

For (1), I just mean that you have a union of two disjoint open sets, so by definition that's disconnected.