Help me understand how we define closed sets in real analysis I'm currently reading the second version of Understanding Analysis by Stephen Abbott and on page $90$ theorem $3.2.8.$ states that "A set $F \subseteq \mathbb{R}$ is closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$". According to definition $3.2.7.$ on the same page "A set $F \subseteq \mathbb{R}$ is closed if it contains its limit points".
The proof was left as an exercise. I think I pretty much proved it but I need help understanding some of the logic involved here. 
If $F$ only contains a single point then there are no sequences in $F$ that do not contain their limit so there are no limit points. Is $F$ still closed? I assume that some sort of logical mambo jambo can be applied here to show that this is the case. To say that $F$ contains all its $0$ limit points does not sit very well with me since we're treating "nothing" as a "something". We're essentially saying that $F$ contains this "nothing". It makes no sense to me. Personally I decided to just embellish the definition with "If $F$ only contains one point then it is closed, otherwise blablabla...". Is this equivalent to the previous definition? 

You are correct when you say that a singleton set does not contain any limit points. 
However, think of the proposition in the following fashion : $F$ is closed, if the set of limit points of $F$ is a subset of $F$.
Then, the set of limit points of a singleton set, is  the empty set, and is therefore contained in $F$. This notion clarifies the doubt of how $F$ can contain nothing, to a better notion of set containment. 
