I'm trying to get a basic definition of completeness in my head (as we are not using the cauchy/limit defn in class). The basic sense that I get, is that a set is complete if there are no 'holes' in it. It makes sence to me that Q is not complete, as I can find some number in $\mathbb{R} \setminus \mathbb{Q}$ (which by defn is not in \mathbb{Q}). I also understand that R is complete because I can always find a '$c$' in $\mathbb{R}$, s.t. '$a$' is less than '$c$' is less than '$b$' (for all $a,b$ in $\mathbb{R}$). What about the interval $(0,2)$ in $\mathbb{R}$? Is this complete? I can always find some $c$ (as noted above) but aren't $0$ and $2$ 'holes' in the set as well?
Edit: sorry, when I used the 'less than' symbol, the site assumed the rest of the post was code and didn't include any of the text beyond the symbol. That is why my post originally abruptly ended mid sentence. I tried using \textless, but my latex is a bit rusty.
Edit 2: we are not given a defn for completeness, only a completeness axiom. - apologies, I cannot reply to comments on my phone.
Edit 3: we are not given the cauchy defn or any defns with limits.
Edit 4: by the comments, I get the impression that an interval like $(0,2)$ is not complete in $\mathbb{R}$, because $0$ and $2$ are 'holes'. What about $[0,2]$ in $\mathbb{R}$? This has no 'holes.' is it complete?
Edit 5: we were told that the supremum is the smallest upper bound. That is, in $(0,2)$ we would say $2$ is the supremum, even though it is not a member of the set. Ie that the supremum can be greater than an element in the set.
I think when you say complete you mean that $S\subseteq \mathbb{R}$ is complete if any subset $A\subseteq S$ bounded from above\below has a supremum\infimum. In the case of $\mathbb{R}$, it is equivalent to the Cauchy sequence definition.
By this definition you can see that a subset $S\subseteq \mathbb{R}$ is complete if and only if it is closed in $\mathbb{R}$. For example $(0,2)$ is bounded but has no supremum, therefore is not complete. This is because for any partial limit, you can find a monotone sub-sequence converging to it.
By the way, when I say closed, I mean in the sense that given a bounded sequence $a_n\in S$, any partial limit is in $S$ when considered regularly in $\mathbb{R}$.