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As discussed in the following question it's impossible to have a pointwise limit of continuous functions converge to $1_{\mathbb{Q}\cap[0,1]}$.

That said, why doesn't the following sequence of functions converge to $1_{\mathbb{Q}\cap[0,1]}$?

Let $S = \{q_j\}_{j=1}^{\infty}$ be an $\mathbb{N}$-indexed sequence of rationals in $[0,1]$.

$f^{(1)}(x) = 1 - \left((x-q_1)\right)^{(1/1)}$

$f^{(2)}(x) = 1 - \left((x-q_1)(x-q_2)\right)^{(1/4)}$

$f^{(3)}(x) = 1 - \left((x-q_1)(x-q_2)(x-q_3)\right)^{(1/9)}$

In General:

$f^{(i)}(x) = 1 - \left(\prod_{j=1}^{\infty}(x-q_j)\right)^{\frac{1}{i^2}}$

If the power $\frac{1}{i^2}$ isn't small enough, then just assume that it's some sufficiently small value like $\frac{1}{e^i}$. The idea is that when $x \in \mathbb{Q}$, $f(x) = 1$, since the term $(x-q)$ will be in the product, and when $x$ is irrational, the $i^2$ root will have the product approach 1.

Nico
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  • I assume you want absolute values inside the product -- you probably don't want to be taking roots of negative numbers. – BallBoy Apr 24 '18 at 22:26
  • Do you want $\prod_{j=1}^i$ instead of $\prod_{j=1}^\infty$ in the definition of $f^{(i)}$? – angryavian Apr 24 '18 at 22:27
  • I'm not convinced that you can choose a power small enough. For each $x$, you can find a subsequence of ${q_j}$ such that $|x-q_{j_k}|<\varepsilon$ for any given $\varepsilon>0$; you have to argue that you can find a power small enough to counteract the effect of multiplying so many $\varepsilon$s in. – BallBoy Apr 24 '18 at 22:29

1 Answers1

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Nothing can be said about limit of $a_n^{b_n}$ if $a_n \to 0$ and $b_n \to \infty$ or $a_n \to 0$ $b_n \to 0$. $f(x)$ is not well defined for $x$ irrational.

Kavi Rama Murthy
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