another proof of the non-existence of a continuous function that switches the rational and the irrational numbers
Suppose is such a function. Since is countable, and are also countable. Therefore the image of is countable. If is not a constant function, then by the intermediate value theorem the image of contains a nonempty interval, so the image of is uncountable. We have just shown that this isn’t the case, so there must be some such that for all . Therefore and . Obviously no number is both rational and irrational, so no such exists.
|Title||another proof of the non-existence of a continuous function that switches the rational and the irrational numbers|
|Date of creation||2013-03-22 16:23:54|
|Last modified on||2013-03-22 16:23:54|
|Last modified by||neapol1s (9480)|