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A set $A$ is said to have least upper bound property if every subset $A_0 \subset A$ has a least upper bound. $\mathbb{R}$ has least upper bound property is well known. Now consider the subset $A = [0,1)$ of $\mathbb{R}$. Does $A$ have the least upper bound property?

I am considering the subset $A_0 = [\frac{1}{2} , 1)$. Clearly it is bounded above in $\mathbb{R}$ with least upper bound $1$. Is $A_0$ is bounded above in $A$? Shall we consider $1$ as its least upper bound?

Thank you for your help.

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You haven't quite got the definition correct: an ordered set $A$ is said to have the least upper bound property if for every non-empty subset $A_0\subset A$ with an upper bound in $A$ has a least upper bound in $A$. By the definition you've written, $\Bbb R$ would not have the least upper bound property, since the empty set is vacuously bounded but has no least upper bound, and since an upper bound of $\Bbb R$ in $\Bbb R$ would be a greatest element (and no such element exists).

As for your example, $A_0$ does not have an upper bound in $A$, so has no least upper bound in $A$. You are correct that it has a least upper bound in $\Bbb R,$ though.

Here's an approach you can take to show that $A$ does have the least upper bound property. Take any non-empty subset $A_0\subseteq A$ such that there is some $x\in A$ such that $a\le x$ for all $a\in A_0$ (that is, such that $A_0$ has an upper bound in $A$). It follows that $A_0$ has an upper bound in $\Bbb R$, so has a least upper bound, say $y\in\Bbb R$. Show that $y\in A,$ and that $y$ is the least upper bound of $A_0$ in $A$. The latter should be easy. For the former, you can use the fact that $A_0$ is non-empty, that $x$ is an upper bound of $A_0$, that $y$ is an upper bound of $A_0$ with $y\le x$ and that $A$ is an interval.

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  • $\begingroup$ Thank you very much for your explanation. It happened because of my misunderstanding. It is clear now. $\endgroup$ – Supriyo Sep 22 '13 at 1:31

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