Discussion
Let X = [0, 1) × [0, 1), Y = [0, 1] × (0, 1) and Z = [0, 1) × (0, 1] be subspaces of the Euclidean space R2. Then :
Answer: C
X is homeomorphic to Z but not to Y.
Z is the image of X under the reflection \((x, y) \mapsto (x, 1 - y)\), which sends [0, 1) to (0, 1] in the second coordinate. A reflection is a homeomorphism of the plane, so X and Z are homeomorphic.
X and Y differ in their edges. The edge points of X (where it is not locally like an open disc) form the two segments \(\{0\} \times [0, 1)\) and \([0, 1) \times \{0\}\), which meet at the corner, a connected set. The edge points of Y form \(\{0\} \times (0, 1)\) and \(\{1\} \times (0, 1)\), two disjoint segments. A homeomorphism carries edge points to edge points, so it would have to carry a connected set onto a disconnected one, which is impossible. Hence X is not homeomorphic to Y.
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