mjhjmtu

This is a problem from my real analysis homework. We are learning the countable sets, and have yet to reach uncountable sets.

Let $f$ be a real function defined on $[0, 1]$. There exists a constant $M$, such that for each finite $n$, and $0 le x_1 < x_2 < cdots < x_n le 1$, we have

$$
|f(x_1) + f(x_2) + cdots + f(x_n)| le M.
$$

Prove $E stackreltextdef= x in [0, 1] : f(x) ne 0$ is countable.

My attempt

I split $E$ into two parts,

$$
beginalign
E^+ &stackreltextdef= x in [0, 1] : f(x) > 0 \
E^- &stackreltextdef= x in [0, 1] : f(x) < 0
endalign
$$

A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, but I didn't gain much from her hint.

Another path I have taken is letting

$$
beginalign
E' &stackreltextdef= x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing \
E'' &stackreltextdef= E setminus E'
endalign
$$

I can prove $E'$ is countable, but $E''$ is still tricky even though it contains less or equal elements than $E$.

Proof by contradiction didn't yield any significant result, either.

A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...

Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.

For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).

In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.

Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.