2022-07-03

Does $f\left(\left[0,1\right]\right)$ must always contain an irrational number?

Aryanna Caldwell

Expert

$\left[0,1\right]$ is uncountable, and so because $f$ is injective, so will $f\left(\left[0,1\right]\right)$. If $f\left(\left[0,1\right]\right)$ did not contain an irrational number, then it would lie in $\mathbb{Q}$, and so be countable. But that would give us a contradiction.

Because $\left[0,1\right]\setminus \mathbb{Q}$ has the same cardinality as $\left[0,1\right]$, there is in fact a bijection from $\left[0,1\right]$ onto $\left[0,1\right]\setminus \mathbb{Q}$.

rjawbreakerca

Expert

where $\left\{x\right\}$ is the fractional part of $x$, though strictly speaking this is a bijection $\left[0,1\right)\to \left[0,1\right)\mathrm{\setminus }\mathbb{Q}$

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