# Assume $\mathbb{R}$ has the least upper bound property. Show that $[0,1] = \{ x\mid 0 \leq x \leq 1\}$ has the least upper bound property

Assume $\mathbb{R}$ has the least upper bound property. Show that $[0,1] = \{ x\mid 0 \leq x \leq 1\}$ has the least upper bound property

My Attempted Proof:

Since $[0,1] \subset \mathbb{R}$, $[0,1]$ is an ordered set. Now take a non-empty $A_0 \subset [0,1]$ such that $A_0$ is bounded above. Since $A_0 \subset [0,1]$, and $\mathbb{R}$ is a complete field, $\exists \alpha \in [0,1]$ such that $\alpha$ is an upper bound of $A_0$.

Corresponding to this $\alpha$ take $\alpha ' \in [0,1] < \alpha$.

Now if $\alpha'$ is not an upper bound of $A_0$, then $\alpha = \sup A_0$ and we are done.

However if $\alpha '$ is an upper bound of $A_0$, put $\alpha = \alpha '$ and take a new $\alpha' < \alpha$ (here we are establishing an inductive argument) and repeat until $\alpha '$ is not an upper bound of $A_0$. Then $\alpha = \sup A_0$ and $\alpha \in [0,1]$.

Thus any non-empty $A_0 \subset [0,1] \subset \mathbb{R}$ that is bounded above, has a least upper bound, completing the proof. $\square$

Is my proof valid and correct? If so how rigorous is it? Any comments and criticism is appreciated. (If however the proof is nonsense, please say so)

You claim that you took an arbitrary set $A_0\subset [0,1]$, and then you define $\alpha$ as some upper bound of $A_0$.

So, at this point:

• $A_0$ is an arbitrary subset of $[0,1]$.
• $\alpha$ is an arbitrarry upper bound of $A_0$.
• $\alpha'$ is an arbitrary element of $[0,1]$, smaller than $\alpha$.

You then claim that

If $\alpha'$ is not an upper bound of $A_0$, then $\alpha=\sup A_0$

which is a false statement. For example, I could have $A_0=[0,\frac12]$, and I could have $\alpha=\frac34$, and $\alpha' = \frac14$. Then, $\alpha'$ is not an upper bound of $A_0$, but $\alpha$ is not the supremum of $A_0$.

Another point where your proof is wrong is that you claim to use an "inductive" argument, but you:

1. Are not making an inductive argument, as there is no clear point at which you are making an inductive step
2. You implicitly imply, by saying "and repeat until $\alpha'$ is not an upper bound of $A_0$", that the process you are describing will stop. There is no reason to believe that, and the process may well go on infinitely long...

• Set $\alpha$ not to "some upper bound" of $\alpha_0$, but to the supremum.
• Show that $\alpha$ must be an element of $[0,1]$.
• Correct me if I'm wrong, but if $\alpha' = \frac{1}{4}$ and $A_0 = [0, \frac{1}{2}]$, then how can $\alpha'$ be an upper bound of $A_0$ if $\alpha' \not\geq \frac{1}{2} \in A_0$? – Perturbative Nov 21 '16 at 12:38
• @Perturbative Sorry, typo. I meant to say that $\alpha'$ is not the upper bound of $A_0$, and yet $\alpha$ is not the supremum of $A_0$. – 5xum Nov 21 '16 at 12:40
• Just another quick question the supremum must exist by the completeness of $\mathbb{R}$? If so here all we need to show is that the supremum must be an element of $[0,1]$, (as you've stated above) correct? – Perturbative Nov 21 '16 at 12:49
• @Perturbative The supremum must exist because the question says Assume $\mathbb R$ has the least upper bound property. – 5xum Nov 21 '16 at 12:50