In each case below say whether the given statement is true for whcih universe
$(0,1)={{(x\in R: 0<x<1})}$
$[0,1]={{(x\in R: 0\le x \le1})}$
$\exists y(\forall x( x>y)$
This means there exists an real number that all other number are greater than it. This is probably true I think in the closed [0,1] universe because zero can be this number. But then zero cant be greater than zero.
$\exists y (\forall x(x\ge y)$
This mean there exist a number that every number is greater or equal to it. I think it true in close [0,1] universe because zero is this number.
The first is false in both universes, because zero does not belong to the universe $(0, 1)$, and for any other choice of a real number $y\in (0, 1)$, there will always be some $x\in (0, 1)$ such that $x\lt y$. If we look at the universe of all reals in $[0, 1]$, $y = 0$ would be a candidate for the existence of a real number less than all others in the universe, except when $x=0$: $x=0$ cannot be strictly greater than $y = 0$. Hence the first proposition is false in either universe.
The second is indeed true in $[0, 1]$: Put $y = 0 \in [0, 1]$. Then every $x\in [0, 1]$, including $x = 0$, is greater than or equal to $y=0$. (The second statement is not true in $(0, 1)$, for the same reason the first isn't true in $(0, 1))$.