(a) The three sets in the question are all of different order types, but the question is whether you can construct a homeomorphism of their topologies regardless of the order itself. If we remove a point from each of two homeomorphic spaces they will remain homeomorphic. If one space is connected so is the other. Remove 1 from (0,1], it is still connected, but any point removed from (0,1) makes it disconnected. So that (0,1] is not homeomorphic to (0,1) (for similar reason [0,1) is not homeomorphic to (0,1) — we need it later, neither is [0,1]). The only two points that being removed keep [0,1] connected are 0 and 1, but since (0,1) is not homeomorphic to [0,1) or (0,1], we conclude that (0,1] is not homeomorphic to [0,1].(b) $X=[0,1]$ , $Y=(0,1)$ , $f(x)=x/2+1/4$ , $g(y)=y$ .(c) Removing 0 makes $\mathbb{R}$ disconnected but removing any point from $\mathbb{R}^n$ leaves it connected.