I want to find a topological embedding $f : X rightarrow Y$ and $g: Y rightarrow X$, yet $X$ is not...
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This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?
general-topology connectedness
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edited Dec 10 '18 at 3:46
user1101010
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asked Sep 24 '13 at 23:15
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