Writing proofs to show that a set is open, not closed and infinite

Question

I have this problem with my homework

6. Let $E$ be a non-empty bounded set of real numbers and put $\alpha = \sup E ,$ and $\beta = \inf E$ . Assume that $\alpha \notin E$ and $\beta \notin E .$ Which of the following statements is true and which is false. In each case justify your answer.

(a) $E$ is an open set.

(b) $E$ is not a closed set.

(c) $E$ is an infinite set.

(d) $( \alpha , \beta ) \subset E$

It's about general topology and inf and sup. We have to state whether each statement is true or false and justify by giving nice proofs. My assumptions are: 1)False 2)True 3)True 4)False

Please, I need help in writing appropriate proofs. I'm new to this course and still don't know how to write proofs. Any help is appreciated. I only have problems proving part d), if anyone can help me. Thank you!

Answer 1

You should do your homework yourself. But you are right about your decision which is true and false.

If the statement is false, you can give just a simple counterexample.

Hints:

b) Assume $E$ is closed. The limit of each convergent sequence in E is in E. Now, use the property of either infimum or supremum to get a contradition.

c) Assume $E$ is finite. What is the supremum and infimum of a finite set?

Answer 2

The convention is to show that a statement is false by providing a counter-example and to show that it's true by giving a general proof.

$(a)$ Consider the counter-example $E=(0,1]\cup[2,3).\alpha=3,\beta=0$ but $E$ is not open.

$(b)$ A closed set contains all its limit points. Show that $\alpha$ is a limit point of $E$ by noting that $N_\epsilon(\alpha)\cap E-\{\alpha\}\ne\phi$. Since $\alpha\notin E,E$ is not closed.

$(c)$ If $E$ were to be finite, it would be closed. We have just shown that $E$ is not closed.

$(d)$ The same counter-example as in $(a)$ works here too.