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Is it possible to construct a bijection from $[0,1]$ to $[0,1]$ with odd number of points of discontinuity ?

Thanks in advance

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  • $\begingroup$ you can basically solve this by drawing a very simple graph of what you want to construct and then setting a function to be invertible and have the graph which you would like. $\endgroup$ – user27182 Mar 21 '13 at 14:46
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Yes. Think about graphs; can you draw a bijection with only one discontinuity? As an example, $$f(x)=\begin{cases}x &:\ x\in[0,1/2)\\ 3/2-x &:x\in[1/2,1]\end{cases}.$$

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Yes. For example, $$f(x)=\begin{cases}\frac{1}{2} & x=0 \\ 1 & x = \frac{1}{2}\\0 & x=1 \\ x &\text{otherwise}\end{cases}$$

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Just draw the graph of $y=x$. Choose any odd no. of points in $[0,1]$ and exchange their values. You are done.

Another solution will be $T$ at an angle of 45 degrees with upper half of its stand deleted.

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