Set of discontinues is always separable and what guarantees existence of $n_1$?

Question

Let $a<b$ be real numbers. We say that a function $f:[a,b]\to\mathbb{R}$ is Baire $1$ if there exists a sequence of continuous functions $(f_n)_{n=1}^\infty$ that converges to $f$ pointwise.

Proposition: If $f:[a,b]\to\mathbb{R}$ is continuous everywhere except possibly countably many points, then $f$ is Baire $1.$

The following proof is provided by my lecturer.

Proof: Let $D_f$ and $C_f$ be set of discontinuities and continuities respectively. By assumption, $D_f$ is countable and $C_f$ is uncountable. Choose a countable dense subset $E_f$ of $C_f.$ Then $D_f\cup E_f$ is countable. List $D_f\cup E_f$ as $$s_1,s_2,...,s_n,....$$ We include $a$ and $b$ in the list. So it becomes $$a,b,s_1,s_2,...,s_n,...$$

We are going to construct $f_n.$ For $n=1,$ find large enough $n_1$ such that for any number $x$ in the list $$a,b,s_1,s_2,...,s_{n_1}$$ there is some $y$ in this list with $$|x-y|<\frac{1}{2}.$$ Let $f_1$ be a piecewise linear function on $[a,b]$ with respect to $$s_1,s_2,...,s_{n_1}.$$ For $n=k,$ choose large enough $n_k$ such that for any number $x$ in the list $$s_1,s_2,...,s_{n_1},...,s_{n_2},...,s_{n_k}$$ there is some $y$ in the list such that $$|x-y|<\frac{1}{2^k}.$$ Let $f_k$ be a piecewise linear function with respect to $$s_1,s_2,...,s_{n_k}.$$

The remaining proof tried to show that $(f_n)$ converges to $f$ pointwise.

Questions: $(1)$ Why can we choose a countable dense subset $E_f?$ Is it because $[a,b]$ is separable, and being a subset of separable in a metric space is also separable. If I want to construct $E_f$ by hand, what should I do?

$(2)$ Why does there exist $n_1$ such that for any $x$ in $$a,b,s_1,s_2,...,s_{n_1}$$ there is some $y$ in the same list with $|x-y|<\frac{1}{2^1}?$

$(3)$ What is an idea of construction continuous $f_n$? In particular, I do not see its 'picture'. For example, if $f$ has finitely many discontinuities, then for each of them, say $x_i,$ consider $(x_i-\epsilon,x_i+\epsilon)$ such that it does not contain other discontinuities. Then define $f_n$ to be a 'peak' function on the open interval and $f_n = f$ outside each of those open interval.

Answer 1

(1) Why can we choose a countable dense subset $E_f$? Is it because $[a,b]$ is separable, and being a subset of separable in a metric space is also separable. If I want to construct $E_f$ by hand, what should I do?

So, in this case the set $C_f$ is uncountable (in fact, co-countable.) This entails that for every open interval $I$ with rational end-points, the intersection $I \cap C_f$ is non-empty. Since the intervals with rational end-points form a countable base for the topology, if we simply select a single point from each of the intersections, the result will be countable and dense.

The process I've out-lined above, would be how one constructs the set $E_f$ "by hand."

(2) Why does there exist $n_1$ such that for any $x$ in $a,b,s_1,s_2,...,s_{n_1}$ there is some $y$ in the same list with $\vert x - y \vert < \frac{1}{2^1}$?

Since the interval $[a,b]$ is compact; the open cover consisting of the open-balls centered at each $s_k$ having radius $\frac{1}{2^1}$, must have a finite sub-cover. Say, this sub-cover consists of the open-balls with centers, $s_{m_1}, s_{m_2}, \ldots, s_{m_j}$ then, taking $n_1 = \max \{ m_1, m_2, \ldots, m_j \}$ we have an integer which meets the stated requirements.

(3) What is an idea of construction continuous $f_n$? In particular, I do not see its 'picture'. For example, if $f$ has finitely many discontinuities, then for each of them, say $x_i$, consider $(x_i−\epsilon ,x_i−\epsilon)$ such that it does not contain other discontinuities. Then define $f_n$ to be a 'peak' function on the open interval and $f_n = f$ outside each of those open interval.

The functions $f_n$ are definable from only a finite fragment of $f$ (in particular it's values at the corresponding points $s_1, \ldots, s_k$.) This is a desirable property.

Answer 2

Separable metric spaces are among the class of hereditarily separable spaces: A subspace of a separable metric space is separable.

(I). A separable metric space is second-countable: It has a countable base (basis). Let $(X,d)$ be a metric space and let $F$ be a dense subset of $X.$ Then $B=\{B_d(f,q):q\in \Bbb Q^+\}$ is a base.

Proof that $B$ is a base for $X$:

For open $U\subset X$ and $x\in U,$ take $r>0$ with $B_d(x,r)\subset U.$ Take $f\in F$ with $d(x,f)<r/3$. Take $q\in \Bbb Q^+\cap (r/3,r/2).$ Let $V(x)= B_d( f,q).$

Then $x\in V(x)\subset B_d(x,r)\subset U$ and $V(x)\in B.$ Therefore $U=\cup_{x\in U}V(x)$ is a union of members of $B.$

If $F$ is countable then (obviously) $B$ is countable.

(II). Any second-countable space X is hereditarily second-countable, for if $B$ is a countable base for $X$ and $Y\subset X$ then $\{b\cap Y:b\in B\}$ is a countable base for $Y.$

(III). Any second-countable space $X$ is separable: If $B$ is a countable base for $X$ then for each non-empty $b\in B$ let $G(b)\in b.$ Any non-empty open $U\subset X$ has a non-empty subset $b\in B,$ so $G(b)\in U.$ Therefore $\{G(b):b\in B\}$ is dense in $X.$

(IV).Hence for a metric space $X$ and for $Y\subset X$ we have $ (X$ separable) $ \implies (X$ 2nd-countable)$ \implies (Y$ 2nd-countable$) \implies (Y $separable).

(V). Instead of all that, a direct method for constructing a countable dense subset of $Y\subset \Bbb R$ is to choose $G(p,q)\in Y\cap (p-q,p+q)$ for $p \in \Bbb Q$ and $q\in \Bbb Q^+$ whenever $Y\cap (p-q,p+q)\ne \emptyset.$ Then
$\{G(p,q): Y\cap (p-q,p+q)\ne \emptyset \}$ is a countable dense subset of $Y.$

(VI). The method of (V) generalizes: For metric space $(X,d),$ if $F$ is a countable dense subset of $X$ and $Y\subset X,$ let $A=\{(f,q): f\in F \land q\in \Bbb Q^+\land Y\cap B_d(f,q)\ne \emptyset\}.$

Choose $G(f,q)\in Y\cap B_d(f,q)$ for each $(f,q)\in A.$ Then $\{G(f,q): (f,q)\in A\}$ is a dense subset of $Y.$

Because if $y\in Y$ and $r>0$ take $f\in F$ with $d(f,y)<r/3.$ (Note: $f$ exists because $F$ is dense in $X )$ and take $q\in \Bbb Q^+\cap (r/3,r/2).$

Now $(f,q)\in A$ because $y\in Y\cap B_d(f,q).$..... And we have $d(y,G(f,q))\leq d(y,f)+d(f,G(f,q))\leq r/3+q<r. $