$f:\mathbb{R}\rightarrow \mathbb{R}$ 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
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2 Answers
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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.
answered Nov 30, 2012 at 13:41
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$\begingroup$ 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. $\endgroup$ Nov 30, 2012 at 14:19
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$\begingroup$ I don;t quite get how you arrive the answer would you mind give more details? $\endgroup$ Nov 30, 2012 at 15:34
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2$\begingroup$ @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. $\endgroup$ Nov 30, 2012 at 19:13
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$\begingroup$ o i didn't notic the absolute value taken on $f$. $\endgroup$ Dec 1, 2012 at 2:55
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$\left\{x:f(x)=0\right\}\\=\cap_{n=1}^\infty\left\{x:|f(x)|\leq\frac{1}{n}\right\}\\=\cap_{n=1}^\infty\left(\mathbb R- \left\{x:|f(x)|>\frac{1}{n}\right\}\right)\\=\mathbb R-\cup_{n=1}^\infty \left\{x:|f(x)|>\frac{1}{n}\right\},\text{an uncountable set.}$
