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The intuition is that $\forall x \in L$ either $x$ is an isolated point or $x$ belongs to $L$ with some interval around it (and I even seem to have managed to prove it). Since both open and closed intervals as well as isolated points are Borel sets, the original claim follows (along with some additional considerations, since there can be uncountably many such intervals/points).

So, more rigorously, we might have three cases for each point $x_0 \in L$ (where, for simplicity, I'm only considering the right vicinity of $x_0$, the left vicinity is analogous):

  1. $x_0$ is an isolated point in $L$:

$$ \exists \epsilon > 0 : \forall x \in (x_0, x_0 + \epsilon) : \nexists \lim_i f_i(x). $$

  1. There is an interval around $x$ in $L$:

$$ \exists \epsilon > 0 : \forall x \in (x_0, x_0 + \epsilon) : \exists \lim_i f_i(x). $$

  1. Neither is true, and there always are points both from $L$ and from $L^C$ arbitrarily close to $x_0$, which is for technical reasons easier expressed as:

$$ \begin{eqnarray} \forall \epsilon > 0 :\ &\exists& x_d \in (x_0, x_0 + \epsilon) : \nexists \lim_i f_i(x_d) \nonumber \\ &\exists& x_c \in (x_0, x_d) : \exists \lim_i f_i(x_c) \nonumber \end{eqnarray} $$

After that some very technical $\epsilon-\delta$ acrobatics allowed me to show the third case leads to a contradiction with the $f_i$ being continuous.

So, is this a correct proof sketch? Didn't I make any mistakes in the proposition for the third case?

As a bonus question, since I'm no topology expert, is there a common name for the third case?


This is an exercise from a book (p. 97, ex. 147, in Russian) on set theory, and the only thing I'm seemingly supposed to know is that a Borel set is an element of the minimal $\sigma$-algebra containing all intervals of the form $[a, b] \mid a, b \in \mathbb{R}$.

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Your third case is entirely possible, and should not lead to a contradiction. For example, if $f_i(x) = i x \sin(1/x)$ with $f_i(0)=0$, $0$ is in the third case.

You can use the definition of Cauchy sequence to express $L$ as a Borel set, in fact a countable intersection of countable unions of closed sets:

$x \in L \iff \forall n\; \exists m\; \forall i,j > m, \; |f_i(x) - f_j(x)| \le 1/n$

Thus $$L = \bigcap_{n=1}^\infty \bigcup_{m=1}^\infty \bigcap_{i,j>m} \{x: |f_i(x) - f_j(x)| \le 1/n \} $$ where since $f_i$ are continuous, $\{x: |f_i(x) - f_j(x)| \le 1/n \}$ is closed.

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  • $\begingroup$ Even if I weren't mistaken in my third case, this is way simpler and more straightforward! Thanks! $\endgroup$ – 0xd34df00d May 7 '18 at 19:42

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