Question

Let A={1,1/2,1/4,1/8,…} and B={ 1/2,3/4,7/8,…}. Explain why supA=supB=1?

Answer 1

Let $A = \{1, 1/2, 1/4, 1/8, \ldots\}$ and $B = \{1/2, 3/4, 7/8, \ldots\}$ . Explain why $\sup A = \sup B = 1$ ?

Solution.

By the definition $\sup A$ is the least number $s$ such that $\mathbf{a}_n \leq s$ for any $\mathbf{a}_n \in A$ . Obviously, value $s = 1$ satisfies this condition.

Consider now the sequence $B = \{b_n\}$ , here $b_n = (2^n - 1) / 2^n$ , $n = 1, 2, \ldots$ and suppose that $\sup B = s < 1$ . Note that $b_n$ increases, $b_n < 1$ and $\lim_{n \to \infty} b_n = 1$ . It means that any $\varepsilon$ -vicinity of 1 contains elements of this sequence. Let's take, for example, $\varepsilon = 1 - s$ and suppose that the terms $b_n$ belong to this vicinity. But then it follows that $b_n > 1 - \varepsilon = 1 - (1 - s) = s$ and by this reason $\sup B$ can't be less than 1.

Thus, $\sup A = \sup B = 1$