Proof that a bounded above set contains a strictly increasing subsequence

Question

If we let $A \subset R$ be bounded above and state that $\sup A \not\in A$, I want to prove that there exists a strictly increasing subsequence in $A$. Here is an attempt I have made but I am not sure if it is correct:

We can pick some $a_1 \in A$, then declare recursively that $a_{n+1} = \frac{a_n + \sup A}{2}$ where by definition $\sup A > a$ $\forall a \in A$, and so is $a_{n+1}$ as it is $< \frac{\sup A + \sup A}{2} = \sup A$. From this we can also clearly say that $(a_n)$ is strictly increasing.

I am uncertain as to whether this is an acceptable proof, I feel like it falls down as we cannot guarantee $a_{n+1} \in A$? Any help is appreciated, thanks!

Answer 1

The problem with your approach is that there's no reason for $a_{n+1}$ to be an element of $A$.

Instead, suppose that $A$ is non-empty with $\alpha=\sup A\not\in A$. Since $\alpha-1$ is not an upper bound for $A$, there exists $a_1\in A$ such that $\alpha-1<a_1<\alpha$.

Similarly, since $a_1$ and $\alpha-\frac{1}{2}$ are not upper bounds for $A$, there exists $a_2\in A$ such that $\max\{a_1,\alpha-\frac{1}{2}\}<a_2<\alpha$.

Continuing this process, one can inductively construct a strictly increasing sequence in $A$ which converges to $\alpha$.

Answer 2

Since $\sup A \notin A$, for any $\epsilon>0$ we have $[\sup A-\epsilon , \sup A) \cap A \neq \emptyset $.

Choose any $a_1 \in A$. Given $a_n$ choose some $a_{n+1} \in [{1 \over 2 } (\sup A + a_n ) , \sup A)$.