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