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Suppose that $f_n:[a,b] \rightarrow \Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?

  • No discontinuities
  • At most ten discontinuities
  • At least ten discontinuities
  • Uncountably many discontinuities
  • Countably many discontinuities
  • No jump discontinuities
  • No oscillating discontinuities

My try :

For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $\varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.

For fourth bullet: To disprove this , consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; x \in \Bbb Q \cap [0,1]\\\\0 &\text{otherwise}\end{cases}$$

Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform

For fifth bullet: To disprove this, consider $$f_n(x)=\begin{cases} \frac{1}{n} & \text{if}\; 0<x<\frac{1}{n} \\\\0 &\text{if}\;x=0 \wedge \frac{1}{n} \leq x \leq 1\end{cases}$$

Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=\frac{1}{n}$ but $f$ continuous on $[0,1]$.

This link answers the sixth bullet


Is my arguments correct ? Can I have a hint for others ?

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    $\begingroup$ My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite. $\endgroup$ – zhw. Dec 3 '18 at 18:55
  • $\begingroup$ The first bullet point is exactly about the case where $f_n$ has no discontinuities. $\endgroup$ – Ingix Dec 4 '18 at 8:35
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First bullet point is correct.

Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.

Hint for third: Very similar construction to the 4th, which is correct.

Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.

For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.

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