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# Non-homeomorphic (TIFR 2013 Math Solution problem 22)

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.