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

Question

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. $\square$

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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)

Answer 1

Your proof lacks rigour.

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 $A_0$.

So, your proof is wrong.

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...

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Hints to rewrite your proof:

• Set $\alpha$ not to "some upper bound" of $\alpha_0$, but to the supremum.
• Show that $\alpha$ must be an element of $[0,1]$.