(a) Yes, it is a linear continuum. For $(z,x)<(z',y)$ we have a point between them (consider two cases $z=z',x<y$ and $z<z',x<1$ ). If $A$ is bounded from above by $(z,x)$ then $\pi_{1}(A)$ is bounded from above by $z$ and it has the greatest element $z'$ . Then let $y=\sup\pi_{2}(A\cap(z'\times[0,1)))$ . If $y=1$ then $(z'+1,0)$ is the least upper bound, and if $y<1$ then $(z',y)$ is the least upper bound. (See also the next exercise.) (b) The other way around it does not work: there is no element between, say, $(0,1)$ and $(0,2)$ . Also, it is not connected. (c) It is. If $(x,a)<(y,b)$ then $((x+y)/2,(a+b)/2)$ is between them. If $A$ is bounded from above by $(x,a)$ , then $\pi_{1}(A)$ is bounded from above by $x$ and let $x'=\sup\pi_{1}(A)\le x$ . If $x'\in\pi_{1}(A)$ then let $y=\sup\pi_{2}((x'\times[0,1])\cap A)\in[0,1]$ , and then $(x',y)$ is the least upper bound. If $x'\notin\pi_{1}(A)$ then $(x',0)$ is the least upper bound. (d) Note that in (c) we needed both points $(x',0)$ and $(x',1)$ to be in the space: the first one for the case $x'\notin\pi_{1}(A)$ and the second one for the case $x'\in\pi_{1}(A)$ and $y=1$ . This suggests that the set in (d) is not a linear continuum. Indeed, $\{0\}\times[0,1)$ is bounded from above but does not have the least upper bound.