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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 $E \stackrel{\text{def}}{=} \{x \in [0, 1] : f(x) \ne 0\}$ is countable.

My attempt

I split $E$ into two parts,

$$ \begin{align} E^{+} &\stackrel{\text{def}}{=} \{x \in [0, 1] : f(x) > 0\} \\ E^{-} &\stackrel{\text{def}}{=} \{x \in [0, 1] : f(x) < 0\} \end{align} $$

A classmate suggested considering $f(x) > \frac{1}{n}$ and $f(x) \le \frac{1}{n}$ 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

$$ \begin{align} E' &\stackrel{\text{def}}{=} \{x \in E : \exists \delta_x > 0, (x, x+\delta_x) \cap E = \varnothing \} \\ E'' &\stackrel{\text{def}}{=} E \setminus E' \end{align} $$

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.

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    $\begingroup$ Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of. $\endgroup$ – Wojowu Mar 2 at 9:20
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    $\begingroup$ The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory. $\endgroup$ – user21820 Mar 2 at 10:06
  • $\begingroup$ @Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero." $\endgroup$ – nalzok Mar 2 at 10:52
  • $\begingroup$ Your classmate's suggestion might be slightly better as considering $f(x) > \frac{1}{n}$ and $f(x) \lt -\frac{1}{n}$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements $\endgroup$ – Henry Mar 2 at 13:50
  • $\begingroup$ @Henry I could have misunderstood her point... $\endgroup$ – nalzok Mar 2 at 13:56
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A classmate suggested considering $f(x) > \frac{1}{n}$ and $f(x) \le \frac{1}{n}$ 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 $f(x) > \frac{1}{n}$ in $E^{+}$.

For each $n \in \Bbb N = \{ 1, 2, 3, \ldots \}$ define $$ E^{+}_n = \left\{x \in [0, 1] : f(x) > \frac 1n \right\} $$ 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 $X$.

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