TIFR 2013 Math Solution Discussion
Question:
The sets \([0,1)\) and \((0,1)\) are homeomorphic.
Hint:
Check some topological invariant.
TIFR 2013 Math Solution
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Discussion:
In \([0,1)\), \(0\) seems to be a special point, as compared to \((0,1)\) where every point has equal importance (or non-importance).
If \(f:X\to Y \) is a homeomorphism then for any point \(a\in X\), \(X- \{ a \} \) and \(Y- \{ f(a) \} \) are homeomorphic with the homeomorphism function being restriction of \(f\) to \(X- \{a \} \).
If \(f: [0,1) \to (0,1) \) is a homeomorphism, then choosing to remove \(0\) from \([0,1)\) we get that \( (0,1) \) is homeomorphic to \( (0,1)- \{f(0) \} \).
Whatever be \(f(0) \), we see that \( (0,1)- \{f(0) \} \) is a disconnected set. Whereas, \((0,1)\) is connected. A continuous image of a connected set is always connected. Hence we are forced to conclude that there was no such \(f\) to begin with.
So the statement is False.
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