Show that $[0,1)$ has no maximum, i.e. $\not \exists \max[0,1)$

Question

Show that $[0,1)$ has no maximum, i.e. $\not \exists \max[0,1)$

My Attempted Proof

Assume $\max[0,1)$ exists and put $\alpha = \max[0,1)$. Now $\alpha < 1$ else $\alpha \not \in [0,1)$.

Put $\gamma= 1- \epsilon$ where $0 < \epsilon \leq 1$. Then $\gamma > \alpha$ for small enough $\epsilon$ and $\gamma \in [0,1)$. Reaching a contradiction. $\square$

---

First off is my proof correct? If so how rigorous is it?o I'm looking to improve my proof-writing skills and rigor in my proofs so if possible please heavily criticize my proof techniques and proof writing and use of concepts. Any comments and criticism is greatly appreciated.

Answer 1

Your intuition for the proof is exactly right. To make it perfectly sound, why not explicitly find an element in $[0,1)$ that is greater than $\alpha$. For example, you know that $\alpha<1$. So the midpoint of $\alpha$ and $1$, namely $\beta:=(\alpha+1)/2$ satisfies $\alpha<\beta<1$, which proves that $\alpha$ cannot be the maximum of the set.

Answer 2

This is true for the same reason that there is no "smallest positive real number."

It can also be proven same way, namely the Archimedean property:

For all $r \in \mathbb R$, there exists some $n \in \mathbb N$ so that $n>r$. One could proceed directly from the axioms to show this ($\mathbb N$ is not bounded above.)

Either way, the relevant corollary: For all $\epsilon>0$, there exists some $n \in \mathbb N$ so that $\frac{1}{n}<\epsilon$.

Now the proof: Suppose that $[0,1)$ has a maximal element $r$. Then there exists some $\frac{1}{n}<1-r$, and hence $r<r+\frac{1}{n}<r+1-r=1$, so $r+\frac{1}{n} \in [0,1)$, contradicting the maximality of $r$.

As an aside, when you are really interested in taking "sufficiently small" epsilons, you can instead defer to sufficiently small $\frac{1}{n}$ and use the Archimedean property in order to make rigorous your argument, and "state" exactly what the number is, since we know it will exist.