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Interior Point of 1/n

  1. Nov 11, 2015 #1
    Hi All,

    A simple question but one for which I cannot seem to get the intuition.

    1. The problem statement, all variables and given/known data


    Find the interior point of {1/n : n ∈ ℕ}.

    2. Relevant equations
    N/A

    3. The attempt at a solution

    Let S = {1/n : n ∈ ℕ}, where S ⊆ℝ

    x is an interior point if ∃N(x ; ε), N(x ; ε) ⊆ S.

    My answer: IntS = (0,1)

    But apparently the answer is ∅, which I do not seem to get.

    Would it be that a small epsilon neighbourhood of x contains some element y such that y is not an element of S?

    Any help is appreciated.

    Edit:

    I seemed to have figured it out. For every x for which we try to find the neighbourhood for, any ε > 0 we will have an interval containing irrational numbers which will not be an element of S.

    Thanks
     
    Last edited: Nov 11, 2015
  2. jcsd
  3. Nov 11, 2015 #2

    Krylov

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    I assume that ##N(x;\varepsilon)## is the open ball of radius ##\varepsilon > 0## centered at ##x##?
    Note that your answer could not possibly be correct, because for any set ##S##, in any topological space, it holds that ##\text{int}\,S \subseteq S##.
    Yes indeed. Try to make this precise. Take an arbitrary ##x = \frac{1}{n} \in S## and an arbitrary ##\varepsilon > 0##. Now indentify an element ##y \in \mathbb{R}## such that ##|x - y| < \varepsilon## but ##y \not\in S##. (You can write down an expression for such ##y## explicitly in terms of ##x## and ##\varepsilon##.)
     
  4. Nov 11, 2015 #3

    Krylov

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    Yes, well done!

    Maybe it's also nice to know that a set ##A## in a topological space is called discrete when every point ##x \in A## has a neighborhood intersecting ##A## only in ##\{x\}##. So, ##S## is an example of a discrete set. However, there are sets (also in ##\mathbb{R}## with the usual metric) with empty interior that are not discrete. (And you know them, too, of course.)
     
  5. Nov 11, 2015 #4
    Thanks a lot Krylov.

    N(x; ε) is indeed the ball centre x, radius ε in this notation.

    - (0,1) cannot be an interior point as 0 is actually a boundary point (i.e. ∀ N(0 ; ε), N(0 ; ε) ∧ ℝ \ S ≠ ∅)

    - Let x ∈ S and x = 1/n. Now take any neighbourgood of x, N(x ; ε) = {y ∈ ℝ : |x - y| < ε}.
    But for any n ∈ ℕ and any φ>0, there exists an irrational number i ∈ ℝ \ S, such that n - φ < i < n.
    But then in any neighbourhood of x = 1/n, there exists a number 1/i. But 1/i ∉ S. Therefore there is no interior points of S, and IntS = ∅.
     
  6. Nov 11, 2015 #5
    Thanks!
     
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