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Prove that there isn't a sequence of continuous function on $[0,1]$ that converges pointwise to the function $f$ on $[0,1]$ defined by $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational.

What I've tried:

By contradiction: Suppose that indeed exists one(say $f_n$) and let's take $a$ in $[0,1]\cap \mathbb{Q}$, then $f_n(a)$ approaches to $0$ when $n$ is sufficiently large, say $n\ge N$. Let's take in particular $n=N$. Since $f_N$ is continuous, there exists a neighborhood of $a$, $(a-d,a+d)$, in which $f_N(x)$ is near $f_N(a)$, and thus is near to $0$. Let's take now $b$ irrational in that neighborhood. Then, for $n$ sufficiently large,say $n\ge K$, $f_n(b)$ is near $1$. This should be the contradiction, since $b$ would be at the same time near $0$ and $1$ (suppose I've chosen the right epsilons). The problem is that the $K$ may be larger than the $N$, and so there is no contradiction. If I then try to make the $N$ larger, then I modify the $d$, which modifies the $b$, and therefore there's no sense in taking that particular $b$. I've also tried to apply continuity to $f_k(x)$, but the neighborhood in which $f_k(x)$ is near $f_k(b)$ (which is near $1$) may not have anything in common with the previous neighborhood...

I wish you could help me. Thanks in advance.

Martin Sleziak
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Diego
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  • Your proof needs to bring in some $\epsilon$ and $\delta$-s, otherwise it is pure handwaving. Saying something is "near" is not rigorouos, and cannot be used in a proof like this. Sure, $f_n(b)$ must be near $1$, but since $a$ and $b$ are not the same number, there could still be a continuous function that goes from $(a, 0)$ to $(b,1)$. – 5xum Dec 27 '15 at 16:27
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    I know that the epsilons and deltas should be in the definitive demonstration, but I think it is worthless to treat the problem in a formal way if I haven't already resolved it in informal way( I mean, with intuition). The epsilons should be like 1/8 each one. – Diego Dec 27 '15 at 16:33
  • @user301198: Have you studied the Baire Category Theorem yet? – John Dawkins Dec 27 '15 at 16:42
  • No...., I'm sorry. – Diego Dec 27 '15 at 16:52
  • @user301198 This example is a good one that demonstrates that sometimes, you need the epsilon delta definitions because "intuition" is not enough. – 5xum Dec 27 '15 at 17:57
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    I would like you to resolve the problem the way you say. – Diego Dec 27 '15 at 18:19
  • See 77307, 75192, 15088, 4738, et c., et c... – Eric Towers Dec 30 '15 at 00:44

1 Answers1

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A function $f$ that is the pointwise limit of a sequence of continuous functions (on $[0,1]$, say) is a function of the first Baire class. Such a function must have at least one continuity point. (Indeed, each non-trivial closed subinterval $[a,b]\subset[0,1]$ must contain at least one continuity point of $f$.) See this article . The proof relies on the Baire category theorem; I don't know of a simpler argument.

Your function has no continuity points, so it cannot be the limit of a sequence of continuous functions.

John Dawkins
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