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Every unctble set of R contains a lim pt in R

  1. Oct 14, 2008 #1
    Every unctble set of R contains a lim pt in R.

    I have assumed it has no lim points and reached the conclusion that this uncountable set has uncountable disjoint sets. Where is my contradiction?
     
  2. jcsd
  3. Oct 14, 2008 #2

    Dick

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    You aren't making much sense there. Think about this, R is the union of a countable number of compact sets.
     
  4. Oct 14, 2008 #3
    Thanks Dick. I have not learned anything about compact yet though.

    I have been taking a topology course and I can recognize that after assuming there are no limit points then I can take the Neighborhood around every point so that it doesn't intersect the set A anywhere except at that point itself.

    It is beginning to sound alot like the discrete metric. I know the contradiction is there. How can the set A be uncountable yet at the same time have disjoint points?

    Is there a way to not bring compact or Topology into this though? Maybe using a fact from a section on uncountable, countable and cardinality? This is my only guess.
     
  5. Oct 14, 2008 #4

    Dick

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    Ok, no compactness arguments. Uh, must think. So you know every point x of the uncountable set, call it A, has an open interval that contains only one member of A. I guess that's what you meant by disjoint. You know every interval contains lots of rational numbers, right? Pick one. Doesn't that give you a 1-1 mapping from A to a subset of the rational numbers? And the rationals are countable, correct?
     
  6. Oct 14, 2008 #5
    So you are saying that in each of my neighborhoods around a point x in A there is a rational number. Define a function mapping each x to a rational in the neighborhood. I understand that would prove it...

    but how do we know there is a rational in each neighborhood?

    Why can't there be uncountable many neighborhoods of x that contain no points?

    And would I just say pick any rational in the neighborhood to map to?
     
  7. Oct 14, 2008 #6

    Dick

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    Pick a definite neighborhood for each x (that has only one member of A in it). Then pick a definite rational from that neighborhood.
     
  8. Oct 14, 2008 #7
    Could you help me with the indexing please...

    Define [tex]x_{\alpha}[/tex] for all x in A, [tex]\alpha \in \lambda[/tex]

    Since Q is dense in R, choose a point [tex]j_{\alpha}[/tex] such that

    [tex]x_{\alpha}[/tex] < [tex]j_{\alpha}[/tex] < [tex]x_{\alpha}[/tex] + [tex]\epsilon[/tex]

    for all [tex]\alpha \in \lambda[/tex]

    Define a function f:R -> Q , [tex]f(x_{\alpha}) = j_{\alpha} [/tex]

    which is 1-1, proving that A is countable, a contradiction.
     
  9. Oct 15, 2008 #8

    Dick

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    That's complete gibberish. Just say what you mean in words first.
     
  10. Oct 15, 2008 #9
    Ok, I meant to say that for every x, choose the nieghborhood of x that contains only the point x from A. Since Q is dense in R we can, from this interval, choose a rational j at the midway point from x to (x+e).

    Repeat this process for every x in A, finding a j in Q. This provides a map that is 1-1.

    What do you think this time?
     
  11. Oct 15, 2008 #10

    Dick

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    Much better. But you do have to do one more thing. Show that the neighborhoods can be chosen in such a way that no two of them overlap. And you can't 'choose a rational at the midway point' unless you pick e especially such that x+e/2 is rational. But there's no need for that complication. Just pick ANY rational. And be more clear about the map. It's not 1-1 with Q. It's 1-1 with a subset of Q (the rationals you selected).
     
  12. Oct 15, 2008 #11
    Hopefully what you mean by this is what I got from assuming that there were no limit points. If there are no limit points I can always choose an e such that the neighborhood of each and every x is disjoint from the entire set. Choose these particular e when constructing the neighborhoods used for the mapping.

    Is that right?
     
  13. Oct 15, 2008 #12

    Dick

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    Using the definition of limit point, you can choose the neighborhoods (call them N(x)) such that A intersect N(x)={x}. You also want to show you can choose them such that N(x1) intersect N(x2) is empty if x1 is not equal to x2. That way you won't accidentally pick the same rational corresponding to x1 and x2. Hint: if e(x) is the original epsilon corresponding to a given x, then changing e(x) to e(x)/2 will do it. (Triangle inequality).
     
  14. Oct 15, 2008 #13
    Dick, you might think that this becomes gibberish too. What do you think?

    [tex]Let A be an uncountable subset of $\mathbb{R}$. Assume that A has no limit points. Then for $\alpha \in \Lambda$ every point $p_{\alpha} \in \mathbb{R}$, there exists $N_{\varepsilon}(p_{\alpha})$ such that $A \cap N_{\varepsilon}(p_{\alpha})=\{p_{\alpha}\}$. After choosing the first $\varepsilon$ that corresponds to the first $\alpha$, we can choose each subsequent $\varepsilon$ so that $\varepsilon_{\alpha}=\frac{\varepsilon}{k}$ with $k={1,2,...}$. Then we can rest assured that for any $p_1,p_2 \in A$, $N_{\varepsilon}(p_1) \cup N_{\varepsilon}(p_2) = \emptyset$.

    \
    We can now choose a unique rational number, call it $v_{\alpha}$ from each $N_{\varepsilon}(p_{\alpha})$ and create a one-to-one function mapping $p_{\alpha} \to v_{\alpha}$. This says that A is countable, which is a contradition.[/tex]
     
    Last edited: Oct 15, 2008
  15. Oct 15, 2008 #14
    What is the deal with my tex??
     
  16. Oct 15, 2008 #15

    Dick

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    You are really good at turning your ideas into gobbledegook index notation. But the idea is somewhat wrong. Call your initial choice of epsilon (from the 'no limit points' argument) for each x in A e(x). Call a ball around x of radius r, B(x,r). Pick an x1 and x2 in A. You know x2 is not in B(x1,e(x1)) and x1 is not in B(x2,e(x2)). So e(x1) and e(x2) are less than |x1-x2|. The balls will intersect if e(x1)+e(x2)>|x1-x2|. Can you show B(x1,e(x1)/2) and B(x2,e(x2)/2) DON'T intersect?
     
  17. Oct 15, 2008 #16
    Choose x1,x2 in A. We know that x1 is not in B(x2,e(x2)) for some e. Thus x1 is not in B(x2,e(x2)/2) for some e.

    The reverse can be said for x2.

    From this we can say that e(x1) +e(x2)< |x2-x1| + |x2-x1| < e(x2)/2 + e(x2)/2 ,

    which is a contradiction hopefully! ...?
     
  18. Oct 15, 2008 #17

    Dick

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    You already picked the e(x)'s with the limit points thing. Don't go changing them now. If there is a point y in B(x1,e(x1)/2) AND B(x2,e(x2)/2) then |y-x1|<e(x1)/2 and |y-x2|<e(x2)/2. But we know |x1-x2|>e(x1) and |x1-x2|>e(x2). The triangle inequality says |y-x1|+|y-x2|>=|x1-x2|. Impossible, right? (You are drawing a picture here, right? I am.) So you conclude B(x1,e(x1)/2) and B(x2,e(x2)/2) DO NOT intersect.

    Now you pick a rational r(x) in each B(x,e(x)/2) and construct your mapping.
     
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