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$f_n(x):[0,1]\rightarrow [0,1]$ are continuous and converges to $f(x)$, then

  1. $f$ is continuous.

  2. Convergence is uniform on $[0,1]$

  3. Convergence is uniform on $(0,1)$

  4. None of above statement is true.

Well, for 1 take $f_n(x)=x^n$, $f(x)=0$ for $ x \in [0,1)$ and $f(x)=1$ at $x=1$ hence 1 is false. For 2 I can give the same counter example as of 1. I have no idea about 3. Please help.

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  • $\begingroup$ How do you show that your example works for 2? What your argument is will determine whether there is significant work left to be done to show that it works for 3. $\endgroup$ – Jonas Meyer Dec 14 '12 at 8:03
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HINT:

You can use the same counterexample with a little work for 3 too.

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  • $\begingroup$ should I take $1-x^n$?, then $f(x)=1\forall x\in (0,1)$ $\endgroup$ – Marso Dec 14 '12 at 7:17
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    $\begingroup$ @Kuttus: Here is another practice problem for you. Let $X$ be a set, and let $F\subset X$ be a finite subset of $X$. If $(f_n)$ is a sequence of real valued functions on $X$ converging pointwise to $f:X\to \mathbb R$, and if $(f_n)$ does not converge uniformly on $X$, then restricting to $X\setminus F$, $(f_n|_{X\setminus F})$ does not converge uniformly to $f|_{X\setminus F}$. Looking at the contrapositive, if you have uniform convergence on a set, and convergence at a point $x$ outside that set, then you have uniform convergence on the union of the original set and $\{x\}$. $\endgroup$ – Jonas Meyer Dec 14 '12 at 7:53
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    $\begingroup$ Be a bit more creative: make the point of discontinuity be, say, at $x=1/2$. $\endgroup$ – Robert Israel Dec 14 '12 at 8:31

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