Originally Posted by

**dwsmith** Consider the following subset of the real line:

$\displaystyle Y=[0,1]\cup (2,3)$,

in the subspace topology. In this space, the set [0,1] is open, since it is the intersection of an open set in R with Y. Similar (2,3) is open as a subset of Y; it is even open as a subset of R (I understand everything here). Since [0,1] and (2,3) are complements in Y of each other, we conclude that both [0,1] and (2,3) are closed as subsets of Y.

Y-[0,1] is open so [0,1] is closed.

Y-(2,3) is closed so why is (2,3) closed?