Thread: Prove that this is a connected subset of R^2

Is $\displaystyle \overline{B((1,0),1)} \cup B((-1,0),1)$ a connected subset of $\displaystyle \Re^2$? I'm sure it is, although I can't find a proof. (Note the "overline" denotes the closure).

Let A,B be two connected subsets of a topological space X such that $\displaystyle A \cap \overline{B} \neq \emptyset$. Prove that $\displaystyle A \cup B$ is connected.

Hint: Is it true that the point $\displaystyle (0,0)$ in one of those sets and a limit point of the other?

Certainly I'd agree that $\displaystyle (0,0)$ is in $\displaystyle \overline{B((1,0),1)}$ and is a limit point of $\displaystyle B((-1,0),1)$. This seems to suggest that there may be some "standard result" that I'm unaware of because I don't see how this helps...

Oh! Does this mean that the set is path connected? Intuitively it seems like you can "pass through (0,0)" but it's a bit tricky to prove this formally.

Originally Posted by Amanda1990

Oh! Does this mean that the set is path connected? Intuitively it seems like you can "pass through (0,0)" but it's a bit tricky to prove this formally.

What basic definition of connectness do you have to work with?
Do you know what a separation of a set is?

No, I haven't come across that term before. Basically, my understanding of a connected set X is one that satisfies any one of the following:

1) If {U,V} is an open cover of X such that $\displaystyle U \cap V = \emptyset$ then either $\displaystyle U = \emptyset$ or $\displaystyle V = \emptyset$.

2) The only subsets of X which are both open and closed in X are $\displaystyle \emptyset$ and X.

3) There exists no open cover {U,V} of X composed of non-empty disjoint sets.

4) For any open (resp. closed) cover {U,V} of X composed of non-empty sets we have $\displaystyle U \cap V \neq \emptyset$

Originally Posted by Amanda1990

Let A,B be two connected subsets of a topological space X such that $\displaystyle A \cap \overline{B} \neq \emptyset$. Prove that $\displaystyle A \cup B$ is connected.

Well then you can use any one of those 4 to prove the second part of the OP which then answers the first part.

But surely in this case $\displaystyle A \cap \overline{B} = \emptyset$ and so the conditions in the second part (which I'm
struggling with anyway, but we'll assume I've proved it) don't seem to hold. So we can't use it to prove the first part. (?)

EDIT: Ah, I see...I was thinking with A and B "the wrong way round". I see now how this second part helps.

Originally Posted by Amanda1990

But surely in this case $\displaystyle A \cap \overline{B} = \emptyset$ and so the conditions in the second part (which I'm
struggling with anyway, but we'll assume I've proved it) don't seem to hold. So we can't use it to prove the first part. (?)

EDIT: Ah, I see...I was thinking with A and B "the wrong way round". I see now how this second part helps.

$\displaystyle A \cap \overline{B} = \emptyset$ means that A contains a point or a limit point of B. If it is only an limit point then what does that mean about open sets containing that point?

I don't understand where we're going with the hint...as in I don't know what this means about open sets containing this point and don't see the relevance. What I have found is that $\displaystyle A \cap \overline{B} \neq \emptyset$ implies that $\displaystyle A \cap \overline{B}$ is connected, but clearly this isn't quite what we are after.