1. Distance between sets

Hi, I'm not sure how to prove this statement:

Prove that if $\displaystyle d(A,B)>0$ then $\displaystyle S = A \cup B$ is disconnected.

I'm using this definition of the distance: d(A,b) = inf{|x-y|: x in A, y in B}

Thanks a lot

2. If $\displaystyle A\cap B\neq\emptyset$ then it should be easy enough to show that $\displaystyle d(A,B) = 0$. Can you see a way to do that?

3. Let $\displaystyle \varepsilon = \frac{{D(A,B)}}{3}$.
Now cover each set: $\displaystyle O = \bigcup\limits_{x \in A} {\mathbb{B}(x;\varepsilon )} \;\& \,Q = \bigcup\limits_{x \in B} {\mathbb{B} (x;\varepsilon )}$.
Now you prove that $\displaystyle O \cap Q = \emptyset$ and therefore you have a disconnection.

4. Originally Posted by nimon
If $\displaystyle A\cap B\neq\emptyset$ then it should be easy enough to show that $\displaystyle d(A,B) = 0$. Can you see a way to do that?
Suppose that $\displaystyle A = \left( { - \infty ,1} \right]\;\& \;B = \left\{ 1 \right\} \cup \left[ {2,\infty } \right)$.

Now quite clearly $\displaystyle A \cap B \ne \emptyset \;\& \;D(A,B) = 0$.

BUT $\displaystyle A\cup B$ is not connected.

5. It is true enough that what I said was unhelpful, but I'm afraid I don't understand your rebuttle. You appear to have contradicted the converse of AKTilted's statement. However I hoped to show that

$\displaystyle A\cap B \neq \emptyset \Rightarrow d(A,B) = 0$,

the contrapositive being:

$\displaystyle d(A,B)>0 \Rightarrow A\cap B = \emptyset$

and therefore that $\displaystyle S$ is disjoint. A proper quashing of that reasoning would be letting $\displaystyle A = [0,1)$ and $\displaystyle B = [1,2]$. Woops!

6. What does that have to do with the union being connected or not?
The OP was about the union being disconnect.
You did not address that directly.