Let $A,B \subset \mathbb{R} $ nonempty sets and bounded such that $A \cap B = \emptyset $ then $sup A \ne supB$

Question

Let $A,B \subset \mathbb{R} $ nonempty sets and bounded such that $A \cap B = \emptyset $ then $\sup A \ne \sup B$

I think that this is wrong and my example is:

$$A=\left\{\frac{1}{2n},\:n\in \mathbb{N}\right\},B=\left\{\frac{1}{2n+1},n\in \mathbb{N}\right\}.$$

We got that: $$A \cap B = \emptyset $$ and $$A\in(0,1)$$

$$B\in(0,1),$$

but $$\sup A = \sup B=1.$$

Can any one tell me whats wrong in this example?

Thanks.

Answer 1

The assertion in the title is false. Consider the two sets $$A = [0,1] \cap \mathbb{Q}$$ and $$B = [0,1] \setminus \mathbb{Q}$$

Here $\overline{A} = \overline{B} = [0,1]$ so $\sup A = \sup B = 1$.

Your counterexample is wrong because the sequences are decreasing to $0$, not increasing to $1$. Try replacing them with $$a_n = 1-\frac1{2n}$$ and $$b_n = 1-\frac1{2n+1}$$instead. Your counterexample would work for $\inf$, not for $\sup$.