1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding accumulation points of ##S=\{\frac{1}{n}+\frac{1}{m}\}##

  1. Nov 26, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##S==\{\frac{1}{n}+\frac{1}{m}, n,m \in \mathbb N\}##, find all the accumulation points of ##S## and decide whether ##S## is compact or not.


    The attempt at a solution.

    I've found that the points of the form ##\frac{1}{n}## and ##0## are accumulation points. For the first, given ##ε>0##, take the point ##\frac{1}{n}+\frac{1}{m}## such that ##m>\frac{1}{ε}##.
    Then, ##|\frac{1}{n}-(\frac{1}{n}+\frac{1}{m})|=|\frac{1}{m})|<ε##. To prove ##0## is an accumulation point, given ##ε>0##, take ##n: n>\frac{1}{ε} \implies \frac{1}{n}<ε##.

    How can I prove that these are the only accumulation points of ##S##?

    For compactness, by the Heine Borel theorem, ##S## is compact ##\iff## ##S## is closed and bounded. But ##0 \in \overline S## and clearly ##0## doesn't exist in ##S##, this means ##S## is not closed so ##S## is not compact.
     
  2. jcsd
  3. Nov 26, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Without loss of generality, assume m <= n. If a > 0 is an accumulation point, can you find an upper bound on m?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted