# Contraposition. Statement with intervals

I want to obtain the contrapositive statement to:

$$x\in \bigcap_{n=1}^{\infty} \left] -\frac1{n}, 1+\frac1{n}\right[ \Rightarrow x\in \left[ 0,1 \right]$$

My guess is:

$$x<0 \text{ or } x>1 \Rightarrow x \leq -\frac1{n} \text{ or } x \geq 1+\frac1{n} \text{ for } n\in \mathbb{N}$$

But I'm not 100% sure. Is this correct?

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It is correct. Even more simply, you could say $x\notin[0,1]\Rightarrow x\notin\cap_{n=1}^\infty]-\frac1n,1+\frac1n[$, but there are times when your more detailed version would be more useful. –  Brian M. Scott Oct 3 '11 at 8:42

Let us overview this: $(p\rightarrow q)$ is equivalent to $(\lnot q\rightarrow\lnot p)$.
Here $p$ is $x\in\bigcap (-\frac{1}{n},1+\frac{1}{n})$ and $q$ is $x\in [0,1]$.
• $\lnot q$, which is $\lnot (x\in [0,1])$, simply put: $x\notin [0,1]$ which as you say is $x<0$ or $x>1$
• And $\lnot p$ is to say $x\notin\bigcap (-\frac{1}{n},1+\frac{1}{n})$, which is to say that for some $n\in\mathbb N$ either $x\le -\frac{1}{n}$ or $x\ge 1+\frac{1}{n}$.
So the contrapositive is indeed $$x<0\lor x>1\Rightarrow \exists n\in\mathbb N:x\le-\frac{1}{n}\lor x\ge 1+\frac{1}{n}$$