There is no continuous function such that and .

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# No swapping

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2 thoughts on “No swapping”

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There is no continuous function such that and .

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Suppose such exists.

Then is connected, because is connected and is continuous. But , which is countable.

So is at most countable. But then has only one element (because if is a connected metric space then or ). Then is constant.

A contradiction follows easily from this fact

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A second solution:

Suppose were such a function. Then for all and . Even more, by the mean value theorem we either have for all or for all .

Let . Then, if are such that , then for all , since the last set of inequalities holds for . Letting approach we get for all . Since obviously can not be zero, is bijective; in particular, it induces a bijection between the rational and the irrational numbers.

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