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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.

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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 $$ a_{n+1}>\sup X-\frac{1}{n+1} $$

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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.

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  • $\begingroup$ Would it be a good idea to prove this using peak points? i.e. let there by finitely many peak points and using this idea to show an is strictly increasing? Thanks $\endgroup$ – Eleanor Richards Nov 26 '16 at 16:10
  • $\begingroup$ I'm not sure what a 'peak point' is; do you mean a point at which the sequence transitions from increasing to decreasing? If so, why would you expect there to be only finitely many? There are plenty of sequences which have infinitely many such points but contain an increasing subsequence. Instead, show that if there is no increasing subsequence, then $\sup X$ must be an element of the sequence, contradicting the assumption that it is not in $X$. $\endgroup$ – Nick Peterson Nov 26 '16 at 16:47

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