(a) For $[0,1]$ all subsets are bounded from above (by $1$ , for example), and the least upper bound of a subset of $[0,1]$ in the real line is the same as the one in $[0,1]$ (it has to be $\le1$ , as $1$ is an upper bound for any such subset).

For $[0,1)$ bounded from above are those subsets that have the least upper bound in the real line less than $1$ (if the least upper bound is $1$ , then there are points in the subset that are greater than $1-\epsilon$ for every $\epsilon>0$ , so there is no upper bound of the subset in $[0,1)$ ). Therefore, the same least upper bound is the one in $[0,1)$ .

(b) $A=X\times Y=$ $[0,1]\times[0,1]$ or $[0,1)\times[0,1]$ . Let $B$ be a bounded from above subset of $A$ , and $M$ be the set of all $x$ -coordinates of all points in $B$ . Since $B$ is bounded from above in $A$ , $M$ is bounded from above in $X$ , so let $m$ be its least upper bound. If $m\notin M$ then there are no points in $B$ with $x$ -coordinate equal to $m$ , but for any $z<m$ there is a point in $B$ such that its $x$ -coordinate is greater than $z$ . Therefore, in this case $(m,0)$ is the least upper bound of $B$ . If $m\in M$ then there are some points in $B$ such that their $x$ -coordinate is $m$ , and the set of all their $y$ -coordinates has the least upper bound $n$ in $[0,1]$ (see (a)). Then, $(m,n)$ is the least upper bound of $B$ .

$A=[0,1]\times[0,1)$ does not satisfy the property, as, for example, $B=\{0\}\times[0,1)$ is bounded from above (by $(1,0)$ , for example), but $B$ has no least upper bound (for any $m>0$ , $(m,0)$ is an upper bound of $B$ , but there is no upper bound $(0,y)$ ).