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Non-homeomorphic (TIFR 2013 Math Solution problem 22)

TIFR 2013 Math Solution Discussion


The sets \([0,1)\) and \((0,1)\) are homeomorphic.


Check some topological invariant.

TIFR 2013 Math Solution

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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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