# Thread: Baby Rudin problem 1.5

1. ## Baby Rudin problem 1.5

Hi,

the whole problem and solution is at: Solutions – Baby Rudin – 1.5 (work in progress) « DaFeda&#039;s Blog. I'm sorry for posting an external link, just trying to avoid writing it all again. However, I do not think you even need to go there to answer my question.

In my proof I write;

If $\displaystyle \alpha > \beta$ then $\displaystyle -inf(A) > sup(-A)$. Since $\displaystyle sup(-A)\geq -x$ for all $\displaystyle x\in A$ this would imply that
$\displaystyle -inf(A) > -x$ , which is a contradiction since $\displaystyle -inf(A) \geq -x$ for all $\displaystyle x\in A$.

Can I use $\displaystyle \geq$ and $\displaystyle >$ in such a way? I have three different solutions available to me from different people, so I am not looking for another way of solving this.

Hope someone has the time to take a look! Thanks.

2. Originally Posted by Mollier
Since $\displaystyle sup(-A)\geq -x$ for all $\displaystyle x\in A$ this would imply that $\displaystyle -inf(A) > -x$ , which is a contradiction since $\displaystyle -inf(A) \geq -x$ for all $\displaystyle x\in A$.
I don't see a contradiction here. The known fact that $\displaystyle a\ge b$ (where $\displaystyle a = -\inf(A)$ and $\displaystyle b=-x$) does not preclude that $\displaystyle a>b$. Moreover, the inequality $\displaystyle \inf(A)<x$ (and, correspondingly, $\displaystyle -\inf(A)>-x$) can be strict for all $\displaystyle x\in A$. This happens when $\displaystyle A$ does not have a minimum, e.g., when $\displaystyle A=\{x\in\mathbb{R}\mid x>0\}$.

3. Originally Posted by emakarov
I don't see a contradiction here. The known fact that $\displaystyle a\ge b$ (where $\displaystyle a = -\inf(A)$ and $\displaystyle b=-x$) does not preclude that $\displaystyle a>b$. Moreover, the inequality $\displaystyle \inf(A)<x$ (and, correspondingly, $\displaystyle -\inf(A)>-x$) can be strict for all $\displaystyle x\in A$. This happens when $\displaystyle A$ does not have a minimum, e.g., when $\displaystyle A=\{x\in\mathbb{R}\mid x>0\}$.
My thinking was that since there is some set $\displaystyle A$ such that $\displaystyle -inf(A)=-x$ for all $\displaystyle x \in A$, then saying that $\displaystyle -inf(A)> -x$ would be a contradiction. I will look for a different approach.
Thanks!

4. I think I can use the fact that since $\displaystyle sup(-A) \geq -x$ then $\displaystyle -sup(-A) \leq x$ which is the definition of $\displaystyle inf(A)$ and so $\displaystyle inf(A)=-sup(-A)$.

5. Originally Posted by Mollier
I think I can use the fact that since $\displaystyle sup(-A) \geq -x$ then $\displaystyle -sup(-A) \leq x$ which is the definition of $\displaystyle inf(A)$ and so $\displaystyle inf(A)=-sup(-A)$.
This shows that $\displaystyle -\sup(-A)$ is a lower bound. Now, you must show that it is greatest among all lower bounds. So, suppose there is a lower bound of A that is greater and use this to get an upper bound of -A which is smaller than sup(-A). This contradicts the leastness of sup(-A).

In general, to show something is a sup/inf, you need to check two things: (1) it is a upper/lower bound and (2) it is (nonstrictly) smaller/larger than any particular upper/lower bound.

6. Here is a different approach.

Suppose that $\displaystyle \lambda =\inf(A)$, $\displaystyle \left( {\forall a \in A} \right)\left[ { - a \leqslant - \lambda } \right]$.
Therefore, there exist $\displaystyle \gamma=\sup(-A)$. Moreover $\displaystyle \gamma \le -\lambda.$.

So suppose that $\displaystyle \gamma < -\lambda$.
That means $\displaystyle \lambda < -\gamma$ which implies $\displaystyle \[\left( {\exists a' \in A} \right)\left[ {\lambda \leqslant a' < -\gamma } \right]$.
But that gives a contradiction: $\displaystyle \gamma < -a'$. Do you see what & why?
So $\displaystyle \gamma= -\lambda$

7. Originally Posted by Guy
This shows that $\displaystyle -\sup(-A)$ is a lower bound. Now, you must show that it is greatest among all lower bounds. So, suppose there is a lower bound of A that is greater and use this to get an upper bound of -A which is smaller than sup(-A). This contradicts the leastness of sup(-A).

In general, to show something is a sup/inf, you need to check two things: (1) it is a upper/lower bound and (2) it is (nonstrictly) smaller/larger than any particular upper/lower bound.
Let $\displaystyle \gamma$ be a lower bound of $\displaystyle A$ ($\displaystyle \gamma \leq x$ for all $\displaystyle x\in A$), and let
$\displaystyle \gamma > -sup(-A)$.
Then $\displaystyle x \geq \gamma > -sup(-A)$ and so $\displaystyle -x \leq -\gamma < sup(-A)$ which makes $\displaystyle -\gamma$ an upper bound of $\displaystyle -A$ that is smaller than $\displaystyle sup(-A)$. Contradiction.

Something like this Guy?

As for your solution Plato, I will take a good look at it later today. Gotta run to work!

8. Originally Posted by Plato
That means $\displaystyle \lambda < -\gamma$ which implies $\displaystyle \[\left( {\exists a' \in A} \right)\left[ {\lambda \leqslant a' < -\gamma } \right]$.
So $\displaystyle \gamma= -\lambda$
$\displaystyle \lambda = inf(A)$ and $\displaystyle inf(A) \leq a \; \forall a\in A$.

$\displaystyle -\gamma = -sup(-A)$ and $\displaystyle -sup(-A) \leq a \; \forall a\in A$.

So what you assuming is that,

$\displaystyle inf(A) < -sup(-A)$, but since $\displaystyle a$ $\displaystyle \geq -sup(-A)$ then how can there exists an $\displaystyle a'$ in $\displaystyle A$ such that $\displaystyle inf(A) \leq a' < -sup(-A)$ ?

Thanks!

9. Originally Posted by Mollier
Let $\displaystyle \gamma$ be a lower bound of $\displaystyle A$ ($\displaystyle \gamma \leq x$ for all $\displaystyle x\in A$), and let
$\displaystyle \gamma > -sup(-A)$.
Then $\displaystyle x \geq \gamma > -sup(-A)$ and so $\displaystyle -x \leq -\gamma < sup(-A)$ which makes $\displaystyle -\gamma$ an upper bound of $\displaystyle -A$ that is smaller than $\displaystyle sup(-A)$. Contradiction.

Something like this Guy?

As for your solution Plato, I will take a good look at it later today. Gotta run to work!
Yeah, that should be fine. I would be a little more explicit with the quantifiers, but the idea is there.