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Math Help - Show that if a ∈ R, then: sup{r ∈ Q : r < a} = a .

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    Show that if a ∈ R, then: sup{r ∈ Q : r < a} = a .

    Show that if a ∈ R, then:
    sup{r ∈ Q : r < a} = a .

    Seems like a very simple question, but the r is throwing me off. Do you prove it the same way as a natural number using an epsilon? or how do you prove a supremum with a rational number.
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  2. #2
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    Quote Originally Posted by habsfan31 View Post
    Show that if a ∈ R, then:
    sup{r ∈ Q : r < a} = a .
    We know that \sigma  = \sup \left\{ {r \in \mathbb{Q}:r < a}\right\} exists.
    Also \sigma \le a because a is an upper bound for the set.
    Suppose that \sigma<a then \sigma<\dfrac{a+\sigma}{2}<a.
    That will give a contradication.
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    Thanks, but im confused. How is a contradiction?
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    Do you realize that between any two numbers there is a rational number?
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    Yes.
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  6. #6
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    K i received help from a friend who took the course, hes not sure if hes right though.

    He says that since Q is dense in R, every nonempty interval of R contains a rational.

    From the definition of the set S = sup{r ∈ Q : r < a} it follows a is an upper bound of the set S. For every ε >0, the density of the rationals in R shows there is a rational r in (a - ε, a). So, a - ε < r < a, which shows a = sup S is the least upper bound of S.

    Is this approach correct?
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  7. #7
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    Well of course. That was the whole point.
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