# Isolation Points?

1. Oct 16, 2013

### mathanon

How do you find the interior points of a subset?

I understand that a point is an interior point if there exists an epsilon neighborhood that is in the set, but I don't know how that would work with specific sets. Any hints?

2. Oct 16, 2013

### Jorriss

If you have a set X and a subset of X called Y, and one wants to know if a point p is an interior point of Y, one need only find a neighborhood of p that is contained in Y.

For example, let X = R, and Y = Q (the rationals). Take an arbitrary point p in Q and take an epsilon neighborhood around p which is contained entirely in Q, that is, p is in (p - ε, p + ε). But it is known that every interval contains an irrational number, which contradicts our assumption that the prescribed interval is in Q. Therefore, as p was arbitrary, Q has no interior points.

Without being more specific to your needs, that is the best I can say.

3. Oct 17, 2013

### mathanon

Thank you! That definitely helps!

4. Oct 17, 2013

### HallsofIvy

Staff Emeritus
Note that you titled this "isolation points" but asked about "interior points". An "isolated point" of a set cannot be an interior point.

For example, if A= (0, 1), the set of all x such that 0< x< 1, the interior points are just points in A itself. That is true because:
if x in (0, 1) then 0< x< 1. Let d1= x, d2= 1- x. If d1< d2, the neighborhood (x-d1, x+d1) is a subset of A. If d2< d1, (x-d2, x+ d2) is in A.

If A= [0, 1], the set of all x such that $0\le x\le 1$, the interior points are again the points in (0, 1). That's true because any neighborhood of "0", (-d, d), includes points outside A (negative numbers to -d) and any neighborhood of "1", (1-d, 1+ d), includes points outside A (numbers larger than 1 up to 1+ d) so "0" and "1", while in the set, are not interior[points].

Some other useful words: we say that point, p, is an "exterior" point of set A if and only if it is an interior point of the complement of A. The complement of (0, 1) is $$(-\infty, 0]cup [1, \infty)$$ and the complement of [0, 1] is $$(-\infty, 0)\cup (1, \infty)$$ both of which have $$(-\infty, 0)\cup(1, \infty)$$ as interior points (so that the "exterior" points of both (0, 1) and [0, 1] are $$(-\infty, 0)\cup (1, \infty)$$. The boundary points of a set are all points that are neither "interior points" nor "exterior points" of the set. Here, the boundary points of both (0, 1) and [0, 1] are the points "0" and "1".

The difference is that those boundary points are in [0, 1] and not in (0, 1). We say that (0, 1) containing none of its boundary points, is an "open" set and [0, 1], containing all of its boundary points, is a "closed" set.

Or course, a set may contain some of its boundary points but not all. (0, 1] is an example. Since neither "none of its boundary points" nor "all of its boundary points" is true, such a set is neither open nor closed.

Although it is unusual, it is possible for a set to have NO boundary points. In that case "none" and "all" are the same, such a set is both open and closed.