Boundary of a metric space Does it make sense to talk about boundary of a metric space, not of subset? 
For example, if $E$ is the metric space consisting of the closed interval $ [ 0,1] $ with the usual metric of $\mathbb{R}$, then is $ \{0,1 \}$ the boundary points of $E$?
I kinda think not, since a boundary point of a set A is defined to be the point where any neighborhood of that point contain some point in A and some point not in A.  But in my example above, any neighborhood of the point $0$ and $1$ only contain point in $E$.  
I'm new to topology so can anyone enlighten me about this? thank you very much. 
Update: Thank you guys for all your answer, it really helps alot.  So let me state what I understand about manifold and can you guys correct me if I'm wrong?? 
I don't know much about general manifold but I only know Euclidean manifold thru the text of Munkres's Analysis on Manifold.
When we say a 1-manifold in $\mathbb{R^3}$, we mean a continuous curve in $\mathbb{R^3}$, is that correct?
let this curve be defined by $f:[0,1] \to \mathbb{R^3}$, and $f$ is a homeomorphism, then by what you guys just state the points $ x_1 = f(0) $ would be a boundary point since for any neighborhood $U$ of $x_1$, there is a neighborhood of the point $0$ that is open in $ \mathbb{H^1}= \{x \in \mathbb{R}: x \geq 0 \}$ that maps homeomorphismly to $U$.  
Is that correct? thank you for all your input.  
 A: Your "kinda think not" is correct, but so is your intuition that the unit interval should have the points $0$ and $1$ on its boundary. 
The structure you want to make this idea work is a manifold. It's a metric space plus some other stuff that makes it resemble a part of $\mathbb{R}^n$ without necessarily being embedded there. 
The "manifold with boundary" paragraphs on this wikipedia page 
https://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary 
describe what you need.
A: Clearly if you regard $E$ as embedded in $\mathbb{R}$, then it has a boundary. But from your comments, I assume you are regarding $E$ as not embedded.
In that case, it is trivial that $E$ is open and closed and hence has no boundary (given the usual definitions - there is a good Wikipedia article on the detail).
But if you are interested in the intuitive concept of boundary as "being on the periphery", rather than the usual definition, then you could still define the boundary in terms of the total ordering inherited from $\mathbb{R}$. In other words you take the boundary to be points $x\in E$ such that there do not exist points $x',x''$ with $x'<x<x''$, where $<$ is the ordering inherited from $\mathbb{R}$ (and points where $x',x''$ do exist are interior points). 
Of course, it depends on what your intuitive concept is. If you regard it as separating $E$ from non-$E$, then it is hard to see how $E$ can have a boundary, and so the usual definition gives the result you want.
Note that this approach of using an inherited order does not generalise well to higher dimensions. If you want to generalise the idea of "being on the periphery" to other cases like subsets of $\mathbb{R}^2$, then you are better off using the machinery of manifold with boundary (also dealt with well on Wikipedia). See also the answer by @EthanBolker .
