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Proof for interval/function boundaries.

  1. Aug 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that the image of f(x)=-x2+2x+1 and the semigroup {x-1 : x<0} are not lower bounded.
    (I do not know the exact translation to english for this last expression, I think they mean that there is not lower limit)

    3. The attempt at a solution
    For the case of the semigroup, I thought of something like this (I am just starting with mathematical formal proofs): Suppose (to look for a contradiction) that there is an infimum M, which must exist if the semigroup has a lower bound, because of the continuity of real numbers. Then M<1/x for all x real <0, and M>M' for all M' smaller than the semigroup.
    Now if I multiply x<0 by a number between 0 and 1, I get another real number "y" smaller than "x" that is also real and <0; so that 1/y belongs to the semigroup and 1/y>M, so M is not a lower bound, this contradicts the first proposition, so that M does not exist.

    I have the feeling that there is something wrong with this, specially in the part of multiplying by another number... help!

    For the case of the function I tried to do something similar, first by proving that f(x+1)<f(x) for all x>1/4, but this just prove that the function is monotonically decreasing after x=1/4, it does not prove the absence of a lower limit.
  2. jcsd
  3. Aug 15, 2014 #2


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    There is an easier way: Suppose [itex]M[/itex] is a lower bound. Then [itex]M < 0[/itex], so [itex]1/(2M) < 0[/itex] also.

    The easiest way is to observe that, by definition of multiplicative inverse, the map [itex]x \mapsto x^{-1}[/itex] is a surjection from the strictly negative reals to the strictly negative reals, which don't have a lower bound.

    Suppose [itex]M[/itex] is a lower bound on the image of [itex]f[/itex]. Then [itex]M \leq f(3) = -7 < 0[/itex]. Since [itex]M < 0[/itex] is a lower bound, there should be no real [itex]x[/itex] for which [itex]f(x) = M - 1[/itex]. Is that the case? [Hint: think about discriminants of quadratic equations.]

    (This does rely on the fact that for every positive real [itex]c[/itex] there exists a positive real [itex]x[/itex] such that [itex]x^2 = c[/itex], but that follows from the least upper bound principle using the same argument by which it is shown that there exists an [itex]x \in \mathbb{R}[/itex] such that [itex]x^2 = 2[/itex].)
  4. Aug 15, 2014 #3
    Ok, I really can't understand what you are telling me for the case of the semigroup. Both x and 1/x are negative; intuitively, I can see that because of continuity of real numbers, for a small lxl=x1 so that 1/-x1 belongs to the semigroup, I can take a smaller lxl=x2 as the average (x1+0)/2, so that 1/-x2 also belongs to the semigroup and is smaller (more negative) than 1/-x1. Can you please explain why did you suggest looking at 1/(2M)<0 ?

    For the case of the function, I think I see what you tell me: it must exist x so that f(x)=M-1, proving that is not so hard, anyway I would have to review discriminants and quadratic equations for doing that.
  5. Aug 15, 2014 #4


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    Is [itex](1/(2M))^{-1} = 2M[/itex] less than or greater than [itex]M[/itex] when [itex]M < 0[/itex]?
  6. Aug 15, 2014 #5
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