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With $I \subset \Bbb{R}$, let $\chi_{I}$ denote the indicator function of I. $$\chi_{I}(X)=\begin{cases} 1, & \text{if $x \in I$} \\ 0, & \text{otherwise} \end{cases}$$ for any $k \in \Bbb{N}$ define $g_{2^k+l}(x)=\chi_{\frac{l}{2^k},\frac{l+1}{2^k}}(x)$ for $l \in {0,...,2^k-1}$. Show ${g_n}^{\infty}_{n=1}$ converges at no point in [0,1].

For every $x_0 \in [0,1]$, $\exists \epsilon_0=1$, $\forall N \in \Bbb{N^*}$, $\exists k \in \Bbb{N}$, $l \in {0,1,...,2^k-1}$ satisfying $2^k+l+1> 2^k+l>N$, and $x_0 \in [\frac{l}{2^k},\frac{l+1}{2^k}]$ such that $$|g_{2^k+l+1}(x_0)-g_{2^k+l}(x_0)|=|\chi_{[\frac{l+1}{2^k},\frac{l+2}{2^k}]}(x_0)-\chi_{[\frac{l}{2^k},\frac{l+1}{2^k}]}(x_0)|=1$$

I am not sure how that leads to convergence at no point.

neveryield
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1 Answers1

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Suppose for a contradiction that ${g_n}_{n=1}^{\infty}$ converged at $x_0$ to $K$; then, for $\epsilon = \frac 1 2$, there exists a sufficiently large $N \in \mathbb N ^*$ such that, for all natural $a > N$, $$ \left\lvert g_a(x_0) - K\right\rvert < \epsilon. $$ Now, by the triangle inequality, for all natural $a, b > N$, $$ \left\lvert g_a(x_0) - g_b(x_0) \right\rvert \leq \left\lvert g_a(x_0) - K \right\rvert + \left\lvert K - g_b(x_0) \right\rvert < 2\epsilon = 1; $$ However, selecting $a = 2^k + l$ and $b = 2^k + l + 1$ (sufficiently large, as your proof asserts is always possible), we clearly obtain a contradiction. So ${g_n}_{n=1}^{\infty}$ does not converge anywhere on $[0, 1]$.

hdighfan
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