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Math Help - Set bounded above

  1. #1
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    Set bounded above

    Let A be bounded above, then prove that \sup(c+A)=c+\sup A, c\in \mathbb R.

    Let a\in A, since A is bounded above, there's supremum, hence a\le\sup A\implies a+c\le c+ \sup A then \sup(a+c)\le c+\sup A, I'm on the right track? How do I finish?

    I think I'm wrong.
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  2. #2
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    Here's what you do: First show that \sup(c+A)\leq c+\sup A. You've pretty much done that already. Now you need to show that c+\sup A\leq\sup(c+A).

    There are a lot of ways to do that. One way is by contradiction: Suppose c+\sup A>\sup (c+A). Then \sup A>\sup(c+A)-c, which means that there is some number x\in A with x>\sup (c+A)-c\geq c+x-c=x, a contradiction.
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  3. #3
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    Quote Originally Posted by Connected View Post
    Let A be bounded above, then prove that \sup(c+A)=c+\sup A, c\in \mathbb R.

    Let a\in A, since A is bounded above, there's supremum, hence a\le\sup A\implies a+c\le c+ \sup A then \sup(a+c)\le c+\sup A, I'm on the right track? How do I finish?

    I think I'm wrong.
    How did you go from a + c <= c + sup A to sup(a + c) <= c + sup A?
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  4. #4
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    Quote Originally Posted by hatsoff View Post
    Here's what you do: First show that \sup(c+A)\leq c+\sup A. You've pretty much done that already
    Just a thing: I determined that \sup(c+a)\leq c+\sup A, can I change a for A ?

    The reason I took a\in A is because I didn't want to say A\le \sup A, since A is a set and it's supremum is a scalar, so that doesn't make sense.
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