Prove that, in the usual topology, both $a$ and $b$ are in the boundary of each $(a,b)$ and that no other point is in the boundary. Suppose that $a$ and $b$ are real numbers such that $ a \lt b$.  Prove that, in the usual topology, both $a$ and $b$ are in the boundary of each $(a,b), [a,b], [a,b),$ and $(a, b]$, and that no other point is in the boundary.
I am in an introduction to proofs class and we have just been introductions to the idea of a boundary. 
Our defintion is: Let $(X, \mathfrak T) $ be a topological space and let $ A \subseteq X.$ A point $ x \in X$ is in the boundary of A if every open set containing $ x$ intersects both $A$ and $X-A$.  
Is there any where you would suggest that I start? 
I think from my definition the set of all boundary points for the first set $(a,b)$ would be $[a,b]$ 
 A: Draw a picture of an interval $A$ from $a$ to $b$ sitting on the real number line.  (It won't matter in this case whether the endpoints are part of the interval.)  Remember that open sets in usual topology on $\Bbb{R}$ are unions of open intervals.  In particular, any open set that contains a particular point $x$ must itself contain an open interval around $x$ (and maybe some other open intervals).
There are essentially five types of points $x \in \Bbb{R}$ to consider:
$$
x<a, \quad x=a, \quad a<x<b, \quad x=b, \quad b<x
$$

Not in the boundary.
For $x<a$, notice that you can find an open interval containing $x$ that lies entirely outside of $A$.  (Just make the the radius $r$ less than the distance $a - x$.)  The other exterior case $(b<x)$ is similar.
For $a<x<b$, you can find an open interval $\{y \in \Bbb{R} : \lvert y - x \rvert < \varepsilon \}$ with $\varepsilon$ small enough that it lies entirely inside of $A$.  How small does $\varepsilon$ need to be?

 $ \varepsilon < \min\{x-a, b-x\} \quad$ or, equivalently, $\quad \varepsilon < x-a \;$ and $\; \varepsilon < b-x$


In the boundary.
Finally, for $x=a$ (the $x=b$ case is quite similar), notice that every open set that contains $x$ contains an interval around $x$ which lies outside of $A$ (on the left) and inside of $A$ (on the right).  No matter how small you make the interval, it always intersects points in $A$ and it's complement.
