# Proof for interval/function boundaries.

1. Aug 14, 2014

### paalfis

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. Aug 15, 2014

### pasmith

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

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

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

(This does rely on the fact that for every positive real $c$ there exists a positive real $x$ such that $x^2 = c$, but that follows from the least upper bound principle using the same argument by which it is shown that there exists an $x \in \mathbb{R}$ such that $x^2 = 2$.)

3. Aug 15, 2014

### paalfis

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.

4. Aug 15, 2014

### pasmith

Is $(1/(2M))^{-1} = 2M$ less than or greater than $M$ when $M < 0$?

5. Aug 15, 2014

thanks!