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
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.
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!