Prove that does not exist a sequence of continous functions that converge pointwise to the Dirichlet function $f:[0,1]\to\mathbb{R}$ defined as $f(x) $=0, if x is rational and $f(x) $=1, otherwise.
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Hint: pick (any?) $x\in {\Bbb Q}\cap[0,1]$, note that there are irrational $y$'s arbitrarily close to $x$, and show that for a suitably large $n$, $f_n(x)$ must be simultaneously close to $0$ and to $1$. (Possibly useful: since the interval is compact, the convergence is uniform, and each $f_n$ is uniformly continuous.) – Tad Mar 04 '15 at 01:00
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@Tad The interval being compact does not mean the convergence is uniform. For example, take $x^n$ in $[0,1]$. – Michael Feb 12 '17 at 01:17
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By theorem of Baire, for every Baire-1 function the points of continuity are a comeager $G_δ$ set . Here the set of discontinuities is the entire interval certainly, the set of points of continuity is not comeager. (a comeager set is the intersection of countably many sets with dense interiors.) For more discussion see here.
Deliasaghi
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