How Do You Determine the Infimum and Supremum of Rational and Sequence Sets?

[Hotsuma]

Homework Statement

Find the infimum and supremum of each of the following sets; state whether the infimum and supremum belong to the set E.

<br /> \item 1. ~~~~E={p/q \in \mathbf{Q} | p^2 &lt; 5q^2 \mbox{ and } p,q &gt;0}. \mbox{ Prove your result. }<br /> \item 2. ~~~~E={2-(-1)^n/n^2|n \in \mathbf{N}. \mbox { Just list your answers}.<br />

Homework Equations

sup E - \epsilon &lt; a \leq sup E.

The Attempt at a Solution

<br /> \item 1. ~~~~ \mbox{Here I claim the supremum of E is }\frac{1}{\sqrt{5}}, \mbox{ in E, and the infimum of E is 0, which is not in Q}.<br /> \item \mbox{Proof: Assume p, q greater than 0. Then } p^2 &lt; 5q^2 \Leftrightarrow p &lt; \sqrt{5} q \Leftrightarrow s = sup E = p/q = \frac{1}{\sqrt{5}} \mbox{how do I prove the infimum part. Hmm...}<br /> \item 2. ~~~~ \mbox{inf E = 0, sup E = 2 (assume 0 in N). }<br />

 

[radou]

p/q < √5, not 1/√5.

 

[Hotsuma]

Thanks, it is late. Woo silly mistakes. How do I prove this result, or is that sufficient? I mean, should I just plug in the typical supremum proof for this result?

Thanks by the way

 

[radou]

Yes, you can prove this from the definition of the supremum and by using the fact that between every two distinct real numbers you can find a rational number.

 

[HallsofIvy]

Homework Statement

Find the infimum and supremum of each of the following sets; state whether the infimum and supremum belong to the set E.

<br /> \item 1. ~~~~E={p/q \in \mathbf{Q} | p^2 &lt; 5q^2 \mbox{ and } p,q &gt;0}. \mbox{ Prove your result. }<br /> \item 2. ~~~~E={2-(-1)^n/n^2|n \in \mathbf{N}. \mbox { Just list your answers}.<br />

Homework Equations

sup E - \epsilon &lt; a \leq sup E.

The Attempt at a Solution

<br /> \item 1. ~~~~ \mbox{Here I claim the supremum of E is }\frac{1}{\sqrt{5}}, \mbox{ in E, and the infimum of E is 0, which is not in Q}.

You mean \sqrt{5} not, 1/\sqrt{5}, but, more importantly, \sqrt{5} is NOT in E since it is not a rational number. 0, on the other hand, is in E: take p= 0, q= 1. And it is the infimum because it is a lower bound (there are no negative numbers in E) while for any n, 1/n, is in E (take p= 1, q= n).

\item \mbox{Proof: Assume p, q greater than 0. Then } p^2 &lt; 5q^2 \Leftrightarrow p &lt; \sqrt{5} q \Leftrightarrow s = sup E = p/q = \frac{1}{\sqrt{5}} \mbox{how do I prove the infimum part. Hmm...}<br /> \item 2. ~~~~ \mbox{inf E = 0, sup E = 2 (assume 0 in N). }<br />

You had better not "assume 0 in N" because then you are dividing by 0 when n=0.

When n= 1, 2-(-1)¹/1²= 3 so 2 is not an upper bound on this set.