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I'm thinking of a particular example to show 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.

However, the only answer provided invoked Baire's theorem, which seems like unnecessary machinery. I'm wondering about a solution that only uses the definition of continuity and point-wise convergence.

This is a slightly different question. Is there any consensus on how I should propose further inquiry into the question?

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    $\begingroup$ Definitely provide a link from your second question to your first one in the question body; it would help answerers to see that suggesting "But, Baire's theorem really is the way to do that" isn't helpful as well as avoid duplicated effort. (It's probably worth it to put links both ways, really) $\endgroup$
    – Milo Brandt
    Dec 30 '15 at 6:44
  • $\begingroup$ @MiloBrandt I have added the link to the question, as you recommended. $\endgroup$
    – Martin Sleziak
    Dec 30 '15 at 10:19
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The additional condition make the question a different one, since the answer to the first version does not answer the second. So yes, you should feel free to post your new question (which is not a "repost"):

How to prove ... without using Baire's theorem, relying only on the concepts of pointwise convergence etc.

(I do suspect that the answers you'll get, if any, will have the idea of Baire's theorem presented in disguise.)

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  • $\begingroup$ I went ahead and took your advice, I think the question is slightly different. $\endgroup$
    – Andres Mejia
    Dec 30 '15 at 5:15
  • $\begingroup$ If I at all understood Baire's theorem, then perhaps the so called straightforward answer would not seem like an unnecessary abstraction. I really need to get my general topology together. $\endgroup$
    – Andres Mejia
    Dec 30 '15 at 5:19
  • $\begingroup$ It's not that much of an abstraction to say that $[0,1]$ cannot be written as a countable union of closed sets with empty interior. $\endgroup$
    – user147263
    Dec 30 '15 at 5:21
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Another way to do this might have been: offer a bounty on the first question, and in the text with the bounty explain the different type of solution you seek.

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