Question

Consider the following statements:
(I) \((0,1)\) and \(\mathbb{R}\) are homeomorphic.
(II) \((0,1)\) and \((0,1]\) are homeomorphic.
(III) \((0,1)\) and \([0,1]\) are homeomorphic.
Choose the correct option.

• A. Only (I) and (II) are false
• B. Only (II) and (III) are false
• C. Only (I) and (III) are false
• D. All (I), (II) and (III) are false

Hint:

Use connectedness after removing a point, and compactness, as the topological invariants to test each pair.

Answer 1

The Correct Option is B

Step 1: Statement (I). The map \(f(x)=\tan\!\big(\pi(x-\tfrac12)\big)\) sends \((0,1)\) continuously and bijectively onto \(\mathbb{R}\), with a continuous inverse. So \((0,1)\) and \(\mathbb{R}\) are homeomorphic, and statement (I) is TRUE.

Step 2: Statement (II). Suppose \(h:(0,1)\to(0,1]\) were a homeomorphism. Removing the point \(1\) from \((0,1]\) leaves \((0,1]\setminus\{1\}=(0,1)\), which is connected. A homeomorphism restricts to a homeomorphism between the spaces with one corresponding point removed, so removing \(c=h^{-1}(1)\) from \((0,1)\) must also give a connected space. But removing any interior point \(c\in(0,1)\) leaves \((0,c)\cup(c,1)\), two disjoint nonempty open pieces, which is disconnected. This contradiction shows no such homeomorphism exists, so statement (II) is FALSE.

Step 3: Statement (III). \([0,1]\) is closed and bounded in \(\mathbb{R}\), hence compact by the Heine-Borel theorem. \((0,1)\) is bounded but not closed, hence not compact. Compactness is preserved by homeomorphisms (the continuous image of a compact set is compact), so a compact space cannot be homeomorphic to a non-compact one. Statement (III) is FALSE.

Step 4: Summary: (I) is true, (II) is false, (III) is false, so exactly (II) and (III) are false.

\[\boxed{\text{Option (B): Only (II) and (III) are false}}\]