# Baby Rudin problem 1.5

• Dec 27th 2010, 11:20 PM
Mollier
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 $\alpha > \beta$ then $-inf(A) > sup(-A)$. Since $sup(-A)\geq -x$ for all $x\in A$ this would imply that
$-inf(A) > -x$ , which is a contradiction since $-inf(A) \geq -x$ for all $x\in A$.

Can I use $\geq$ and $>$ 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.
• Dec 28th 2010, 02:38 AM
emakarov
Quote:

Originally Posted by Mollier
Since $sup(-A)\geq -x$ for all $x\in A$ this would imply that $-inf(A) > -x$ , which is a contradiction since $-inf(A) \geq -x$ for all $x\in A$.

I don't see a contradiction here. The known fact that $a\ge b$ (where $a = -\inf(A)$ and $b=-x$) does not preclude that $a>b$. Moreover, the inequality $\inf(A) (and, correspondingly, $-\inf(A)>-x$) can be strict for all $x\in A$. This happens when $A$ does not have a minimum, e.g., when $A=\{x\in\mathbb{R}\mid x>0\}$.
• Dec 28th 2010, 02:57 AM
Mollier
Quote:

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

My thinking was that since there is some set $A$ such that $-inf(A)=-x$ for all $x \in A$, then saying that $-inf(A)> -x$ would be a contradiction. I will look for a different approach.
Thanks!
• Dec 28th 2010, 03:50 AM
Mollier
I think I can use the fact that since $sup(-A) \geq -x$ then $-sup(-A) \leq x$ which is the definition of $inf(A)$ and so $inf(A)=-sup(-A)$.
• Dec 28th 2010, 06:30 AM
Guy
Quote:

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

This shows that $-\sup(-A)$ 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.
• Dec 28th 2010, 07:14 AM
Plato
Here is a different approach.

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

So suppose that $\gamma < -\lambda$.
That means $\lambda < -\gamma$ which implies $\[\left( {\exists a' \in A} \right)\left[ {\lambda \leqslant a' < -\gamma } \right]$.
But that gives a contradiction: $\gamma < -a'$. Do you see what & why?
So $\gamma= -\lambda$
• Dec 28th 2010, 09:16 PM
Mollier
Quote:

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

Something like this Guy?

As for your solution Plato, I will take a good look at it later today. Gotta run to work!
• Dec 28th 2010, 10:49 PM
Mollier
Quote:

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

$\lambda = inf(A)$ and $inf(A) \leq a \; \forall a\in A$.

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

So what you assuming is that,

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

Thanks!
• Dec 29th 2010, 06:11 AM
Guy
Quote:

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