Let $A \subset X$ a set of all characteristic functions of finite sets. Show that no sequence in $A$ converges to the constant map $c(x)=1$. 
Let $X = \mathbb{R}^{\mathbb{R}}$ and $A \subset X$ a set of all characteristic functions of finite sets. Show that no sequence in $A$ converges to the constant map $c(x)=1$.

Any sequence in $A$ is of form $(\chi_{B_1}(x), \chi_{B_2}(x), \dots)$ for some finite $B_i$’s.
Do they mean that a sequence converges to $c(x)=1$ if I can find some $x$ for which $(\chi_{B_1}(x), \chi_{B_2}(x), \dots) =(1,1, \dots)$?
What’s the correct intuition here?
 A: Break it down, using the definitions:
$X$ is

 just the collection of functions from $\mathbb R$ to $\mathbb R.$

A $sequence$ in $X$ is then

a collection $(f_n)_{n\in \mathbb N}:f_n\in X.$

If $f_n\in A\subset X$, then $f_n(x)=0$

for all but finitely many values of $x.$

Therefore $\{x: \exists  n\in \mathbb N( f_n(x)=1)\}$

is countable.

If $f_n\to c$ then for each $x\in \mathbb R,$

 $f_n(x)=1$ if $n$ is large enough,

which is impossible.
edit: it may be easier to see this using the ordered pair definition of the $f_n.$ Each $f_n$ is a set whose members are either $\langle x,0\rangle$ or  $\langle x,1\rangle$ and the number of the latter is finite. Taking the union over $n$, it follows that the total number of elements of the form $\langle x,1\rangle$ in the collection $\{f_n\}_n$ is at most countable. If the $f_n$ were to converge to the constant function $c=1$ then, for each real number $x$ some member of $f_n$ would have to contain an element of the form $\langle x,1\rangle.$ This is a contradiction, since $\mathbb R$ is uncountable.
A: A sequence $(f_n)_n$ in $\Bbb R^{\Bbb R}$ converges to $f$ iff $$\forall x \in \Bbb R: f_n(x) \to f(x)\tag{1}$$
Now, suppose $\chi_{B_n} \to c$. The sequence $\chi_{B_n}(x)$ for any fixed $x \in \Bbb R$ is a sequence of $0$ and $1$'s that converges to $1 = c(x)$ by assumption and fact $(1)$. Do there is some $N_x$ so that $n > N_x$ implies $\chi_{B_n}(x)=1$ or $$\forall n > N_x: x \in B_n$$
Now, we can find some $N$ so that for $C_N:= \{x\mid N_x = N\}$ is uncountable, by the pigeon hole principle (or else $\Bbb R$ would be countable which it's not). Now it follows that $C_N \subseteq B_n$ for all $n >N$, a flagrant contradiction with all $B_n$ being finite.
So no such sequence can exist.
