$d(A,B)=\min\{ d(a,b): a\in A\text{ and }b\in B\}>0$ then $A\cup B$ is not connected

If $$d(A,B)=\min\{ d(a,b): a\in A\text{ and }b\in B\}>0$$ then $$A\cup B$$ is not connected.

We claim that the two sets are disjoint, suppose it is not we have $$d(A,B)=0$$

So now we have to say any union of two arbitrary disjoint sets is not connected

I am stuck here

So there exists $$r>0$$ with $$d(a,b)>2r$$ for all $$a\in A$$, $$b\in B$$. What can you say about the following unions of open balls of radius $$r$$? $$\bigcup_{a\in A}B_r(a)\qquad\text{and}\qquad \bigcup_{b\in B}B_r(b)$$

As A and B are disjoint then you can show that They are actually both open and close in AUB considering the relative topology. Hence they will form a separation of A and B.

Union of two disjoint sets can be connected: $$(0,1)$$ and $$\{0\}$$ are disjoint but their union is connected.

Let $$C=A \cup B$$. If $$C_1=\overline {A} \cap C$$ and $$C_2=\overline {B} \cap C$$ then $$C=C_1\cup C_2$$ and $$C_1,C_2$$ are non-empty disjoint closed subsets of $$C$$, so $$C$$ is not connected. [They are disjoint because $$d(\overline {A} ,\overline {B} )=d(A,B)>0$$].

• In general, if $$d(A,B)>0$$ then $$A$$ and $$B$$ are disjoint sets.

• If $$d(A,B)=0$$ then $$A$$ and $$B$$ may be or may not be disjoint sets

Note that for the closure of a subset $$C$$ of the metric space $$X$$ we have $$\bar{C} = \{ x \in X \ | \ d(x,C)=0\}$$

Back to our problem. Since $$d(A,B)>0$$ we have $$d(a, B)>0$$ for any $$a\in A$$ and so $$A \cap \bar{B} = \emptyset$$. Similarly, $$\bar A \cap B = \emptyset$$.

Now, the closure of $$A$$ as a subspace of $$A\cap B$$ equals $$\bar A \cap(A\cup B) = (\bar A\cap A) \cup (\bar A\cap B)= A$$, and so $$A$$ is a closed subspace of $$A\cup B$$. Similarly $$B$$ is a closed subspace of $$A\cup B$$. Hence, $$A$$, $$B$$ form a separation of $$A\cup B$$, and so $$A\cup B$$ is not connected.