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Let $S = [a,b]\cap\mathbb{Q}$. Two definitions:

  • $\mathcal{X}_S(x) = 1$ if $x \in S$ and $0$ otherwise;
  • $\mathcal{X}_{k}(x)= 1$ on the first $k$ entries of $S$ and $0$ otherwise, $k \in \mathbb{N}$.

Prove that $\mathcal{X}_k \nearrow \mathcal{X}_S$. I take $\nearrow$ to mean that $\mathcal{X}_k$ is increasing up to its limit $\mathcal{X}_S$. This seems intuitively true, but how may I conclusively show it?

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    Do you mean pointwise convergence? it doesn't converge uniformly – 2017-01-30
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    That seems right. – 2017-01-30
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    You have to show that $\chi_k(x) \leq \chi_{k+1}(x)$ for all $k$ and $\lim_{k \to \infty} \chi_k(x) = \chi_S(x)$. Where exactly are you stuck? – 2017-01-30
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    Since the set of rationals is countable, you can enumerate (i.e. create a bijection) every element of $[a,b]\cap \mathbb{Q}$ by $x_1,x_2,\dots$. Since then $\mathbb{Q}$ can be well ordered (by order induced by the bijection) we may assume that $x_1\le x_2\le \cdots$; So given any $x\in S$ you can find some $m$ such that $x_m=x$ and $\chi_n(x)=1$ for all $n\ge m$ – 2017-01-30
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    @user160738 does the last line imply that $\mathcal{X}_k \leqslant \mathcal{X}_{S}$ for all $k$? – 2017-01-30
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    If $q_n$ is the $n$th rational in your enumeration of $\mathbb Q\cap [a,b]$ then $\chi_k(q_n)=0$ for $k$\chi_k(q_n)=1=\chi_S(q_n)$ for $k\geq n.$... And $\chi_k(x)=0=\chi_S(x)$ for every irrational $x\in [a,b].$ – 2017-01-30

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