Set theory proof for $\min C = \bigcap _{\alpha \in C} \alpha$ Let $C$ be a set of ordinals
a) $\sup C$ is the least ordinal $\beta$ such that for all $\alpha\in C$, $a\leq \beta$.
b) If $C$ is transitive (and thereby ordinal), then $\sup C \notin C$ if an only if $\sup C$ is a limit ordinal or $\varnothing$.
c) $\min C = \bigcap _{\alpha \in C} \alpha$
Provide a proof.
I suppose the best starting point would be to show that $\sup C\notin C$ is not true when $C$ is a successor ordinal, correct? This I guess makes sense insofar as the successor ordinal of any given ordinal, by the nature of it being a successor, is included in the set $C$ correct? I'm sure there's a more eloquent way of putting it but that seems to make sense. (Is that adequate?)
For this (c) though I am not even sure what this means. What does this notation mean with $\alpha \in C$ as a subscript? The minimum of $C$ is equal to the intersection of $\alpha$ where $\alpha$ is in C... I'm almost sure I'm reading the notation wrong. 
Any help would be greatly appreciated!
 A: Here are some hints:
a) Recall that $\sup C = \bigcup C$. First of all prove that $\bigcup C$ is an ordinal. 
Having verified that, suppose that $\gamma$ is such that $\alpha \le \gamma$ for all $\alpha \in C$, then consider two cases: If $C$ doesn't have a maximal element, then $\alpha \in \gamma$ for all $\alpha \in C$. Note that in this case $\sup C = \bigcup C \subseteq \gamma$ which yields $\sup C \le \gamma$. If $C$ does have a maximal element $\delta$, then verify that $\sup C = \delta$, which also implies $\sup C \le \gamma$.
b) If $C$ is a successor ordinal, it has a maximal element $\delta$. You verified in a) that $\delta = \sup C \in C$. If $C = \emptyset$, then clearly $\sup C \not \in C$. Finally let $C$ be a limit ordinal. I claim that $\sup C = C$ in this case, which yields $\sup C \not \in C$ - by regularity.
c) First of all prove that $\bigcap C  = \bigcap_{\alpha \in C} \alpha$ is an ordinal.
Now, if $\gamma < \alpha$ for all $\alpha \in C$, then $\gamma \in \bigcap C$ (note that $\bigcap C = \bigcap_{\alpha \in C}$) and hence $\bigcap C \le \min C$. On the other hand: If $\gamma < \bigcap C$, then $\gamma \in \alpha$ for all $\alpha \in C$ and thus $\gamma < \min C$. Therefore $\min C = \bigcap C$.
