Closed sets and finite sets I was wondering if someone could clear this up for me. According to Trench(A real Analysis book), A set is closed if and only if it contains all its limit points. It also further said that all finite sets are closed. But in my mind(I may be wrong), there can be a finite set that will have interior points but not limit points. Which makes it open, not closed. I wonder if I am missing something. 
Again, does the definition of a a closed set tell us that every point is a limit point? I thought some of its points are also boundary points.
thanks
 A: Since you're working out of a real analysis book, let's only talk about the real numbers for this problem. We know that a limit point $x$ of a set $A$ is some point where there is a sequence of members if $A$, call it $\lbrace a_n \rbrace$, where each $a_i$ gets closer and closer to the limit point $x$ as $i \rightarrow \infty$ , i.e. $|x-a_i| < \varepsilon$ for arbitrarily small $\varepsilon$. Notice that these points $a_i$ can $\underline{\text{not}}$ be the same as $x$, also by definition.
With this definition in place, we say a set $A$ is closed if and only if it contains all of it's limit points; e.g. we know the interval $(0,1)$ is not closed because the sequence $\lbrace \frac{1}{n} \rbrace$ converges to $0$, but $0 \not\in (0,1)$. We could also show that $[0,1]$ is closed if we wished. Notice that with these two examples, sequences are easy to construct, because we have lots of numbers between $0$ and $1$ to make sequences with. (note also that we may defined closed as $A^{\prime} \subseteq A$, where $A^{\prime}$ is the set of limit points of $A$; this may come in handy later).
Now let us consider a (vastly) smaller set; $\lbrace 1,2,3,7 \rbrace$. As a perfectly nice subset of $\mathbb{R}$, we can then start to talk about it's openness or closedness with respect to the topology on the $\mathbb{R}$. So, if we wanted to know if $A$ is closed, let's find some limit points! Particularly, let's see if $1$ is a limit point.
Well, if we take a sequence of numbers from this set, we'll notice something strange; all the elements of any $\lbrace a_n \rbrace$ will be either $1,2,3,7$. With this in mind, think of how we can make a sequence that gets closer and closer to $1$ but never touches $1$; in fact, it turns out that we can't, because there are no elements arbitrarily close to $1$ in our set which are not $1$ itself! So, we have that $1$ is not a limit point.
If we extend this to all the other elements, we then see that $A = \lbrace 1,2,3,7 \rbrace$ has no limit points! Then, because it contains all $\underline{\text{NONE}}$ of them (vacuously), we know that $A$ is a closed set. So, what we have found is that some sets don't have any limit points, and because of this that makes them closed. If the bit about vacuous truth is confusing, then we only need to know that $A^{\prime} = \varnothing$, and then clearly $\varnothing \subseteq A$, and $A$ is again closed.
