Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f:\mathbb{R}\rightarrow \mathbb{R}$ is function such that $\forall\epsilon>0$, the set $\{x:|f(x)|>\epsilon\}$ is finite. We need to show $\{x:f(x)=0\}$ is uncountable. Could any one give me hints?

Thank you

share|cite|improve this question
up vote 8 down vote accepted

HINT: For each $n\in\Bbb Z^+$ let $A_n=\left\{x\in\Bbb R:|f(x)|>\frac1n\right\}$. Show that

$$\{x\in\Bbb R:f(x)\ne 0\}=\bigcup_{n\in\Bbb Z^+}A_n\;,$$

so that $\{x\in\Bbb R:f(x)\ne 0\}$ is a countable set.

share|cite|improve this answer
Even $\bigcup A_n$ is countable, but we don't actually know how $f$ is defined, then how to conclude $\{x\in\mathbb{R}:f(x)=0\}$ is uncountable?it may be countable infinite. – Mathematics Nov 30 '12 at 14:19
I don;t quite get how you arrive the answer would you mind give more details? – Mathematics Nov 30 '12 at 15:34
@Mathematics: No, $\{x\in\Bbb R:f(x)=0\}$ can’t be countably infinite. If it were, $\Bbb R$ would be the union of two countable sets, $\{x\in\Bbb R:f(x)=0\}$ and $\{x\in\Bbb R:f(x)\ne 0\}$, and would therefore be countable, but we know that $\Bbb R$ is uncountable. – Brian M. Scott Nov 30 '12 at 19:13
o i didn't notic the absolute value taken on $f$. – Mathematics Dec 1 '12 at 2:55

$\left\{x:f(x)=0\right\}\\=\cap_{n=1}^\infty\left\{x:|f(x)|\leq\frac{1}{n}\right\}\\=\cap_{n=1}^\infty\left(\mathbb R- \left\{x:|f(x)|>\frac{1}{n}\right\}\right)\\=\mathbb R-\cup_{n=1}^\infty \left\{x:|f(x)|>\frac{1}{n}\right\},\text{an uncountable set.}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.