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Math Help - Show that every open interval has infinitely many points

  1. #1
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    Show that every open interval has infinitely many points

    I'm sure there's a better way to do this, but is this logically sound? (thanks for reading)


    1. Let (a,b) be an open interval.

    2. Assume (a,b) has finitely many points.

    3. If (a,b) has finitely many points, then ∃x ∈(a,b) such that x<α ∀α ∈(a,b)
    and ∃y ∈(a,b) such that y > β ∀β ∈(a,b).

    4. Since (a,b) is an open interval, ∀α ∈(a,b) ∃ε such that (α - ε)∈(a,b). Likewise,
    ∀β ∈(a,b) ∃δ such that (β + δ)∈(a,b). Thus, ∃ε such that (x - ε)∈(a,b) and ∃δ such that (y + δ)∈(a,b).

    5. But (x - ε) < x and (y + δ) > y. This contradicts 3., therefore if (a,b) is an open interval then (a,b) has infinitely many points.
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  2. #2
    Senior Member Tinyboss's Avatar
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    I think it's basically sound. You could make it shorter by only getting a contradiction at the left or right end of the interval, instead of both ends. But I think an easier way would be to say that any finite collection of real numbers can be well-ordered consistent with the standard order, so if you take two "adjacent" points a,b then the open interval (a,b) would have to be empty. One way or another, it comes down to the Archimedean property of the reals.
    Last edited by Tinyboss; May 30th 2011 at 09:43 PM.
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  3. #3
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    I was wondering if we might not be able to construct a direct proof (i.e., not a proof by contradiction) using something along the lines that every open subset A (interval) of the real numbers contains a neighborhood Q for each x\in A. In particular, this is true for the endpoints of the interval A = (a, b). Using the Archimedean property, say on b, we can always find another real number y\in Q_b. In other words, \exists\epsilon>0\ s.t.\ b -\epsilon< y < b,\ Q_b = (b-\epsilon, b+\epsilon).

    Additionally, now that I think about it, isn't this part of an important theorem in real analysis? We have to prove that b is not contained in the interval? It's been awhile since I've done analysis, so I'll have to dig through my textbook, but I recall something like this (and the approach I mentioned above) that took a whole class to drill into us (and half the proof showed up on a test).

    Nevermind that, after a quick scroll through my text on continuity, I believe I had the Heine-Borel theorem in mind, which is quite the opposite (S is compact iff S is closed and bounded). We had to prove that b was actually contained in the set.
    Last edited by bryangoodrich; May 30th 2011 at 07:31 PM. Reason: Addendum + Addendum
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  4. #4
    Senior Member Tinyboss's Avatar
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    Or we could just show (or take as given) that if a<b, then there exists some c in (a,b). Then there exists d in (a,c), and there exists e in (a,d), and so on.
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  5. #5
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    Quote Originally Posted by Tinyboss View Post
    Or we could just show (or take as given) that if a<b, then there exists some c in (a,b). Then there exists d in (a,c), and there exists e in (a,d), and so on.
    How would you write that in an appropriate mathematical way that proves the interval is infinite?

    Would induction be the way to go to do this direct proof?
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  6. #6
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    Quote Originally Posted by SunRiseAir View Post
    How would you write that in an appropriate mathematical way that proves the interval is infinite?
    For any two real numbers x<y then x<\frac{x+y}{2}<y..
    Define c_1=\frac{a+b}{2}.
    If n\ge 2, define c_n=\frac{a+c_{n-1}}{2}.
    Can you show that the collection \{c_n\} is infinite and each member belongs to (a,b)~?
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