Is the Set {1/n: n is a Counting Number} Equal to the Interval (0,1]?

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The set P = {1/n: n is a counting number} is not equal to the interval (0,1] because P contains only rational numbers, while (0,1] includes irrational numbers. The discussion clarifies that P is a subset of (0,1], but they are not equivalent sets. Additionally, it is established that an open set in a metric space can exist without containing boundary points, as all points in an open ball are interior points.

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consider the set P={1/n:n is counting number}, my classmate said that P is equal to (0,1] but actually i don't agree with him since (0,1] contains irrational numbers. is he correct? also, is it possible for a set not to contain both interior and boundary points?
 
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Can you elaborate? You're correct, of course, that P = {1/n:n is counting number} contains only rational numbers (and not even all of the rational numbers in the unit interval), and so is not equal to (0,1], but maybe he meant something other than equality of sets?

As for your second question: Are you asking if there exists a set that fails to contain both interior an boundary points? In an open set in a metric space, by definition, all points are interior points, so we could take any open ball, for example: every point is interior, and there are no boundary points.
 

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