Finding Interior Points in a Subset: Hints & Tips

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To find the interior points of a subset, a point is considered an interior point if there exists an epsilon neighborhood around it that is entirely contained within the subset. For example, the set of rational numbers has no interior points because any neighborhood around a rational number will include irrational numbers. In contrast, for the open interval (0, 1), all points within that interval are interior points, while the endpoints 0 and 1 are not. The distinction between open and closed sets is highlighted, with open sets containing none of their boundary points and closed sets containing all of them. Understanding these concepts is crucial for identifying interior, exterior, and boundary points in various sets.
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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?
 
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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.
 
Thank you! That definitely helps!
 
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.
 

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