For a non empty finite set A whose all the elements are integers, its supremum and infimum are also integers.

Question

I am trying to prove the statement. This is my attempt. Can someone help me with it?

Attempt: Proof by contradiction,

Let's say all the elements of A are integers, but supA is not an integer.

Then by approximation property of real numbers, for any $\epsilon$>0, $\exists$ a $\in$ A such that supA- $\epsilon$ < a $\leq$ supA, which tells me that 'a' can be a non-integer, since the choice of $\epsilon$ was arbitrary.

Hence we are done.

Am I doing this right?

Answer 1

Seems right to me, assuming $A$ is finite. You might want to state more precisely what $\epsilon$ is exactly. You could state this in terms of $\sup(A)-\lfloor\sup(A)\rfloor$, for example. If you choose $\epsilon:=\frac{\sup(A)-\lfloor\sup(A)\rfloor}{2}$, you can be sure that $a\in A$ is chosen from an interval that does not intersect $\mathbb{Z}$.

Answer 2

No, it is not correct. How do you jump from the fact that $\varepsilon$ is arbitrary to the conclusion that $a$ can be a non-integer? Where did you use the fact that $\sup A\notin\mathbb Z$?

Answer 3

HINT:

A finite set of numbers will have a largest number and a smallest number. Now, if a set has a largest number, that will be its supremum. So the supremum will be in the set. Same for the infimum. So...