The problem is to prove that the union of open subsets of $\mathbb{R}$ is open.
However, the only definitions that I have to work with are:
A set is closed if it is not equal to the empty set and if contains all of its limit points.
A point $p$ in the set $A$ is a limit point of $A$ if all the points in $A$ converge to $p$.
With those definitions, I have surmised that:
An open set is a non-empty set that does not contain (all of) its limit points.
I also know:
A union of sets is a set that contains all the elements that are in any of the sets in the union.
Below is what I currently have for the proof. Please tell me if this is a sufficient proof, if you see any mistakes, etc.
Proof:
Let $U$ be the set of an arbitrary union of subsets of $\mathbb{R}$.
I.e., $U$ = $\cup$ $X_n$ where $X$ is an arbitrary subset of $\mathbb{R}$ and $n \in \mathbb{R}$.
Since $U$ is comprised of subsets that do not contain their limit points, then $U$, the union of these subsets, does not contain its limit points.
Therefore, $U$ is open.
Is this a sufficient proof?
Best Answer
Once you have a definition of either open or closed sets, usually you define the other type of set by saying that an open set is the complement of a closed set (or equivalently, a closed set is the complement of an open set). So if you accept this axiom, and you have a definition of a closed set, then you need to show that the intersection of two closed sets is closed and then you will have that the union of two open sets is open.