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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)

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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...

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 $[0,1]$.
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  • $\begingroup$ 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 $\alpha' \not\geq \frac{1}{2} \in A_0$? $\endgroup$ – Perturbative Nov 21 '16 at 12:38
  • $\begingroup$ @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 $A_0$. $\endgroup$ – 5xum Nov 21 '16 at 12:40
  • $\begingroup$ 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? $\endgroup$ – Perturbative Nov 21 '16 at 12:49
  • $\begingroup$ @Perturbative The supremum must exist because the question says Assume $\mathbb R$ has the least upper bound property. $\endgroup$ – 5xum Nov 21 '16 at 12:50
  • $\begingroup$ Thanks for your answer yesterday, really appreciated it, I figured out today (thanks to your answer) how the proof should be written and corrected the flaws in my logic $\endgroup$ – Perturbative Nov 22 '16 at 17:33

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