Confusion about the definition of interior points on Rudin's real analysis So the definition of an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N$ is a proper subset of $E$ in $X$(metric space). 
(BTW, A neighborhood of $p$ is a set $Nr(p)$ consisting of all q such that $d(p,q)<r$ for some $r > 0$. )
Suppose I choose $X$ to be $[1,3] \cup \{5\}$, and $E$ is $[1,3]$,if I pick $p =3$ , for $r = 1$, therefore the neighborhood of $p$ within $r$ should be the proper subset of $E$. So point $p$ should be the interior point of $E$, but I know it's not. Am I misunderstanding the concept of metric space(from book's definition it just a set, whose elements are points with distance property), or something else went wrong. Thank you!
 A: $3$ is an interior point.
And $[1,3]$ is an open set.... in this space.
Why do you think $E$ shouldn't be open?  I imagine it is because it has sharp cut off points.   But why does a sharp cut off point mean a set is not open? I imagine it is because you can go past a sharp cut out point and not be in the set anymore.  But if the entire UNIVERSE gets cut off, then you can't go past the cut off point.  So... the sharp cut off point doesn't matter.
Yes, it's "on the edge" but that's okay because the other side of "the edge" (the numbers just larger than $3$) don't exist.  
Think of it this way.  You are snug inside a set if you can go in every direction possible  and have other points around you. Okay, that's way too informal but that is what it essentially means. In $[1,3]$ the point, $2.1$ say, is an interior point because you can go either up or down to get $(2.1-e, 2.1+e)$ and those are all in $[1,3]$.  
And at $3$ you can go down and get $(3-e, 3]$ are all in $[1,3]$.  Now you may argue you can't go up.  And that's because there isn't any "up" to go.  The number $(3, 3+e)$ simply do not exist in the space at all.  So all the numbers within $e$ of $3$ are $(3-e, 3]$.  That is the open neighborhood.  And $3$ IS an interior point.
A third way of thinking of it:  $N(3,e) = \{x\in [1,3]\cup \{5\}| d(x,3)< e\} = \{x\in \mathbb R|d(x,3)< e\} \cap ([1,3]\cup \{5\}) = (3-e, 3+e) \cap ([1,3]\cup \{5\}) = (3-e, 3]$.  That's the open neighborhood around $3$ in this space.
Anyway, $3$ is an interior point.  Just because the universe was sliced by laser beam exactly at the point $3$ and nothing exists on the other side doesn't change anything because..... well, the universe was sliced and the other side doesn't exist in the universe any more.
A: A neighborhood $N$ of $p$ is a set containing $B_X(p,r)=\{x\in X:d(p,x)<r\}$ for some $r>0$.
The concept makes sense when you consider a subset $E$ of a metric space $X$; a point $p\in E$ may or not be interior.
If $E=[1,3]$ as a subset of $\mathbb{R}$ with the usual metric, then $3$ is not an interior point of $E$. Indeed every neighborhood of $3$ in $\mathbb{R}$ contains numbers greater than $3$.
If $E=[1,3]$ as a subset of $X=[1,3]\cup\{5\}$, then $3$ is an interior point; indeed
$$
B_X(3,1)=\{\color{red}{x\in X}:|x-3|<1\}=(2,3]\subset E
$$
In a different way, a singleton is closed in every metric space. Thus $\{5\}$ is closed in $X$ and therefore its complement in $X$, namely $[1,3]$, is open in $X$, so each of its points is interior.
