Definition of closed sets

kimkibun

Good day!

Im currently reading the book of Steven R. Lay's "Analysis with an Introduction to Proof, 3rd ed.". According to his book, if a subset S of ℝ contains all of its boundary then it is closed. But i find this wrong since if we consider S={xεQ;0≤x≤2}, then it can be shown that S contains all of its boundary points (using the fact the Q is dense in ℝ), but it is not closed since the closure of S is the interval [0,2] which is not equal to the set itself. am i correct?

SteveL27

Quote by kimkibun

Good day!

Im currently reading the book of Steven R. Lay's "Analysis with an Introduction to Proof, 3rd ed.". According to his book, if a subset S of ℝ contains all of its boundary then it is closed. But i find this wrong since if we consider S={xεQ;0≤x≤2}, then it can be shown that S contains all of its boundary points (using the fact the Q is dense in ℝ), but it is not closed since the closure of S is the interval [0,2] which is not equal to the set itself. am i correct?

Is √2 a boundry point of S? Is it in S?

HallsofIvy

Because, as you say, Q is dense in the real numbers, every irrational number between 0 and 2 (in fact, every number in S as well) is a boundary point, not just 0 and 2. S is NOT closed because it does not contain the irrational numbers. The closure of S is the interval [0, 2] including all rational and irrational numbers in that interval.

algebrat

May not be relevant, but you should also check which "space" you are in. In the space Q, the closure of S is S. In R, the closure of S is [0,1].

A. Bahat

Elaborating on algebrat's response, "closed subset" is a relative concept, depending on what topological space that subset is embedded in. (Obviously "open subset" is relative to the larger space as well.) This is in contrast to a property like compactness, which is intrinsic.

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HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	Because, as you say, Q is dense in the real numbers, every irrational number between 0 and 2 (in fact, every number in S as well) is a boundary point, not just 0 and 2. S is NOT closed because it does not contain the irrational numbers. The closure of S is the interval [0, 2] including all rational and irrational numbers in that interval.

algebrat	Definition of closed sets May not be relevant, but you should also check which "space" you are in. In the space Q, the closure of S is S. In R, the closure of S is [0,1].

A. Bahat	Elaborating on algebrat's response, "closed subset" is a relative concept, depending on what topological space that subset is embedded in. (Obviously "open subset" is relative to the larger space as well.) This is in contrast to a property like compactness, which is intrinsic.