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gbean
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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?