Prove that $X$ contains a strictly increasing sequence which converges to $\sup X$

Question

Suppose that $X ⊂ \mathbb{R}$ is bounded above and that $\sup X \notin X$. Prove that $X$ contains a strictly increasing sequence which converges to $\sup X$.

I've started by assuming there to be an increasing sequence and using its definition to show that as $m$ tends to infinity, $a_n < l$ for all $n < m$. This satisfies first criteria of suprenum but not sure where to go next? Thanks.

Answer 1

You can't assume what you want to prove.

Hint: suppose you have already chosen $a_1<a_2<\dots<a_n$ in $X$ such that $a_k>\sup X-\frac{1}{n}$. Show you can choose $a_{n+1}>a_n$ with $$ a_{n+1}>\sup X-\frac{1}{n+1} $$

Answer 2

Assuming you have a sequence to begin with doesn't seem like a good way to go -- you need to CONSTRUCT a sequence that satisfies the given properties.

Focus on the definition of supremum as least upper bound. Show that for each $n\in\mathbb{N}$, there must exist $a_n\in X$ so that $\lvert a_n-\sup X\rvert<\frac{1}{n}$. Clearly, $a_n$ defined this way converges to the supremum.

Now, there's no guarantees that $(a_n)$ is an increasing sequence. But, we can find a subsequence which does what we want. Prove that $(a_n)$ contains a strictly increasing subsequence. How? Consider the fact that $X$ doesn't contain its supremum.