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#### issacnewton

##### Member

- Jan 30, 2012

- 61

Let A be an infinite subset of \(\mathbb{R} \) that is bounded above and let \( u= \mbox{sup }A \). Show that there exists an increasing sequence \( (x_n) \)with \( x_n\in A\) for all \(n\in\mathbb{N} \) such that \(u=\lim (x_n)\).

My try. Since A is infinite, it means \(A\neq \varnothing\). We can consider two cases here. Case 1 is when \( u\in A\). So we can construct a

constant sequence \( x_n=u \) which converges to u. Case 2 is when \(u \neq A\). Since \( A\neq \varnothing\), let \( x_1 \in A\). So we have

\( x_1 < u \). Since \( u= \mbox{sup }A \), there exists \( x_2\in A\) such that \(x_1 <x_2 \). Again \( x_2 < u \). So we can go on building the sequence. So consider the set \( \{x_n |n\in\mathbb{N} \} \). This is bounded below by \(x_1 \) and bounded above by \(u\) and its increasing, so by monotone convergence theorem, this is convergent and \( \lim(x_n)=\mbox{sup }\{x_n |n\in\mathbb{N} \} \). I think my reasoning is correct till this point. Now I need to prove that \( u=\lim (x_n)\). So I basically need to prove that

\[ \mbox{sup }\{x_n |n\in\mathbb{N} \}=u \]

I could prove that \( \forall n\in \mathbb{N} (x_n \leqslant u) \) using the way the sequence is constructed. But I am having trouble with proving that

if \( k < u \) there should exist some \( n_1 \in\mathbb{N} \) such that \( k< x_{n_1} \).

I have seen some proofs floating on the net. But didn't quite understand it. So wanted to post here.