Open or Closed: Analyzing the Intersection of Rationals and an Interval

[gbean]

Homework Statement

Given the set S = the intersection of the rationals and the interval [0, 1], is S open or closed?

Homework Equations

Definition of open: for all elements of S, there exists epsilon > 0 such that the neighborhood (x, delta) is a subset of S.

The Attempt at a Solution

Since the rationals are dense in the reals, then I will have an infinite union of open neighborhoods around the rational elements of S, which should mean that the set is open.

But this is wrong because of the following explanation I was given:
Taking some neighborhood around an element of S, there exists a rational element and an irrational element in this neighborhood.

Since there exists an irrational element in this neighborhood, then this neighborhood is not an element of S, and S is not open.

Also, S is not closed. The complement of S is not open because there exists a rational element in the neighborhood of an element of S, so the neighborhood is not a subset of S. S is not open, so S is closed.

I basically don't understand the explanation for why the set is neither open nor closed. If there exists an irrational element within the neighborhood around a rational element of the set, does this mean that the neighborhood is no longer a part of the set? So a neighborhood can only be contained within a set if every element within that neighborhood is also an element of the set?

 

[gbean]

But then I can take an infinite union of open sets (neighborhoods) that will only contain the rational elements, so shouldn't this be open?

 

[HallsofIvy]

For a set to be open, every member must be an "interior point". That means there must be some neighborhood of the point that itself contains only points from the set. Here, that would mean there must be some interval around a rational number that contains only rational numbers. But there exist irrational numbers in every interval so that is not true.

For a set to be closed, its complement must be open. The complement of this set is "all numbers less than 0" union "all number greater than 1" union "the irrational numbers between 0 and 1". Again, it is true that there exist rational numbers in any interval so there cannot exist an interval around an irrational number between 0 and 1 that contains only irrational numbers. That set is not open so the set of rationals between 0 and 1 is not closed.

 

[HallsofIvy]

But then I can take an infinite union of open sets (neighborhoods) that will only contain the rational elements, so shouldn't this be open?

No, you can't. NO neighborhood (open interval) contains only rational numbers, much less an infinite union.

 

[gbean]

Ok, that makes sense. Thank you!