Verification. Find the supremum and infimum of each set.

Question

Looking for quick verification and explanation if wrong.

$E = \{x \in \mathbb{R} \space \big| \frac{1 +(-1)^n}{n}, n \in \mathbb{N}\}$

$n = 1$ $\to x = 0$

$n = 2$ $\to x = 1$

$n = 3$ $\to x = 0$

$n = 4$ $\to x = \frac{1}{2}$

$n = 5$ $\to x = 0$

$n = 6$ $\to x = \frac{1}{3}$

$n= 8$ $\to x = \frac{1}{4}$

$x \in [0,1] \forall x \in E$ therefore $\sup(E) = 1$ and $\inf(E)=0$

$E = \{x \in \mathbb{R}\ :\ x^2 - 2x + 3 \gt x^2 \space \space \text{and} \space \space x \gt 0\}$

In this set it is easy to notice that in order for $E \gt x^2$ then we need to find all values where $-2x+3 \gt 0$ then $x^2 - 2x +3$ will always be greater than $x^2$

$-2x \gt -3$

$x \lt \frac{3}{2}$

means $x \lt 1.5$

The set of all values that satisfy $E$ fall into the interval of $(0, \frac{3}{2})$

$\sup(E) = \frac{3}{2}$ and $\inf(E) = 0$

$E = \{x \in \mathbb{R} : \frac{1}{1 + (-1)^n}, \space n \in \mathbb{N}\}$

$x = 2 \to \frac{1}{2}$

$x = 3 \to DNE$

$x = 4 \to \frac{1}{2}$

The is only one possible number are $\frac{1}{2}$

$\inf(E) = \sup(E) = \frac{1}{2}$

$E= \{\frac{p}{q} \in \mathbb{Q}\ :\ p^2 \lt 5q^2 \text{ and } p,q \gt 0\}$

about all I can come up with is:

$\frac{p}{q} \lt \sqrt{5}$

which I feel like has infinite answers for $\sup(E)$ and $\inf(E)$

Answer 1

For the last part the supernum is $\sqrt 5$ and the infimum is $0$

Your answers to the other parts are correct.

Answer 2

You cannot find the supremum and infimum by just writing down some points in the set and then making a guess. You have to give a rigorous proof of your calim.

In the first case $E=\{\frac 2 {n}: n \text { even} \} \cup \{0\}$. Can you show from this that $0\leq x \leq 1$ for all $x \in E$. Once you do this you wiil get $\sup E=1 $ and $\inf E=0$ by what you have done in your answer.

Your answer to the second part is right.

The third question is wrong since $E$ is not well defined set.

For the fourth question the infimum is $0$ and the supremum is $\sqrt 5$. Use approximation of irrationals by rationls to prove this.