what is the boundary of this set?
{1\n : n $\displaystyle \in$ N}
is the bd = (0, 1)?
question, does every set have a boundary point?
Let's assume that the set of real numbers is the space.
Use the convention that $\displaystyle \mathbb{N} = \left\{ {1,2,3, \cdots } \right\}$.
Define $\displaystyle \mathbb{F} = \left\{ {\frac{1}{n}:n \in \mathbb{N}} \right\}$.
The boundary of the set $\displaystyle \mathbb{F}$ is $\displaystyle \mathbb{F}\cup \{0\}$.
Well, I think maybe Plato's answer should be explained a little more. Clearly any point of $\displaystyle \xi\in\mathbb{F}=\left\{\frac{1}{n}:n\in\mathbb{N} \right\}$ is an element of $\displaystyle \partial\mathbb{F}$. Why? Well because $\displaystyle \xi$ itself is in $\displaystyle \mathbb{F}$ and you can find a neighborhood around it such that no other point of $\displaystyle \mathbb{F}$ is contained in that neighborhood (why is that? distance between any two points of $\displaystyle \mathbb{F}$ is ____ which means ____). Thus every point $\displaystyle \xi$ of $\displaystyle \mathbb{F}$ has a neighborhood surrounding it such that (except for $\displaystyle \xi$) is contained in $\displaystyle \mathbb{R}-\mathbb{F}$. So why does that imply that $\displaystyle \xi\in\partial\mathbb{F}$? Once you see that you can deduce that $\displaystyle \mathbb{F}\subset\partial\mathbb{F}$. Now the reason why $\displaystyle 0\in\partial\mathbb{F}$ is because $\displaystyle 0$ is a limit point (but not an interior point)...why does that make it a boundary point?
But it do not take that much to see it. Not if we use definitions.
A point is a boundary point of a set if each open containing the point contains a point in the set and a point not in the set.
Because each point of $\displaystyle \mathbb{F}$ is rational then any open set containing a point of $\displaystyle \mathbb{F}$ also contains a point not in $\displaystyle \mathbb{F}$.
Now $\displaystyle 0 \notin\mathbb{F}$ but is a limit point so it is a boundary point.
You must learn the definition and what it means.
$\displaystyle x$ is a boundary point of a set $\displaystyle A$ if and only if any open set containing $\displaystyle x$
must also contain a point in $\displaystyle A$ and a point not in $\displaystyle A$.
Think what means if $\displaystyle A=\mathbb{R}$.
well as far as what I think* I know, the reals are infinite.
So I would have to say that you wouldnt be able to get a point outside of the reals that is +-epsilon.
thats my best guess, but unless you can use infinity as a boundary? can you?
or do i need to do some reading because I am way off?
Some text use the following definitions for metric spaces:
"interior point" has the usual definition: x is an interior point of set A if and only if there exist some neighborhood $\displaystyle N_\epsilon(x)$ which is a subset of A.
"exterior point": x is an exterior point of set A if and only if it is an interior point of the complement of A.
"boundary point": x is a boundary point of set A if it is neither an interior point nor an exterior point of A.
"open set": A is an open set if it contains none of its boundary points.
"closed set": A is a closed set if it contains all of its boundary points.
It is obvious how a set may be neither open nor closed- if it contains some but not all of its boundary points.
But in order to be both open and closed, it must contain all and [/b]none[/b] of its boundary points. That is only possible if "all= none"- that is that it has NO boundary points.
Of course, one of the basic properties of any topological space (not just metric spaces) is that the entire set and the empty set must be both closed and open. In terms of these definitions, all points are interior points of the entire set, so the entire set has NO exterior points and NO boundary points. Conversely, all points are exterior points of the empty set. The empty set has no interior points and no boundary points. For $\displaystyle R$ or any "connected set" the only sets that are both open and closed (sometimes called "clopen") are the entire set and the empty set.
But suppose X= $\displaystyle [0,1]\cup [2,3]$ with the metric d(x,y)= |x-y|. Here, the only points we are considering are points in the two intervals- there do not exist any other points. Then the neighborhood $\displaystyle N_{0.1}(3)$ is (2.9, 3], NOT "(2.9, 3.1)" because the points form 3 to 3.1 are not in our space- for our purposes, they "do not exist". Here, 0 and 1 are interior points of [0,1] and 2 and 3 are interior points of [2,3]. The set of all interior points of [0,1] is [0,1] itself, the exterior points of [0,1] are [2,3] and it has no boundary points. [0,1] is both open and closed in this topology, as is [2,3].