Statement is $\exists I = (x_o - \frac{1}{n}, x_o + \frac{1}{n}), n \in \mathbb N$, s.t $f(x) > 0 $ $\forall x \in I$
When negating the part after "such that", would it be
$f(x) \leq 0$ $\forall x \in I $ or $\exists x \in I$ s.t $f(x) \leq 0$ ?
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3It would be the second. To negate "$\forall x$, property p is satisfied" means that "$\exists x$ such that property p is not satisfied" – Julian Mejia May 13 '19 at 02:40
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oh no. now my proof is completely invalid :( – MinYoung Kim May 13 '19 at 02:40
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BTW the negation of the entire statement would be "$\forall n\in\mathbb{N}$, there is some $x_n\in (x_0-1/n,x_0+1/n)$ s.t. $f(x_n)\leq 0$". This negation would imply (if your function is continuous) that $f(x_0)\leq 0$ since $x_n\to x_0$. (And this would be a contradiction if $f(x_0)>0$) – Julian Mejia May 13 '19 at 02:53
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what about the part where $\forall I = (x_o - \frac{1}{n}, x_o + \frac{1}{n})$? – MinYoung Kim May 13 '19 at 03:05
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$I$ depends on $n$, so this $I$ is actually indexed by $n$, $I_n=(x_0-1/n,x_0+1/n)$. So, "$\forall I$,$ \exists x\in I$ with $f(x)\leq 0$" is the same as "$\forall n$, $\exists x_n\in I_n$ with $f(x_n)\leq 0$ – Julian Mejia May 13 '19 at 03:12
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oh right got it. – MinYoung Kim May 13 '19 at 03:41