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- Apr 14, 2013

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I want to determine the following sets:

- $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$

- $\displaystyle{\bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$

- $\displaystyle{\bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )}$

I have done the following:

- Let $\displaystyle{x\in \bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$. This means\begin{align*}&\forall n\in \mathbb{N} \ : \ x\in \left (-\frac{1}{n},\frac{1}{n}\right ) \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ -\frac{1}{n} < x< \frac{1}{n} \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ -1 < nx< 1\end{align*}

To be able to divide by $x$ we have to consider three cases: $x=0$, $x>0$, $x<0$.

Case 1: $x=0$

We get \begin{align*}& \forall n\in \mathbb{N} \ : \ -1 < n\cdot 0< 1 \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ -1 < 0< 1\end{align*} which is true for every $1\leq n\in \mathbb{N}$. .

Case 2: $x>0$

We get \begin{align*}&\forall n\in \mathbb{N} \ : \ -1 < nx< 1 \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ -\frac{1}{x} < n< \frac{1}{x}\end{align*}

Since $\mathbb{N}$ is unbounded, we can find a value for $n$ that is bigger than $\frac{1}{x}$. So this inequality is not true for every $1\leq n\mathbb{N}$. This means that if $x>0$ the element $x$ cannot belong to the intersection $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$.

Case 3: $x<0$

We get \begin{align*}&\forall n\in \mathbb{N} \ : \ -1 < nx< 1 \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ \frac{1}{x} < n< -\frac{1}{x}\end{align*}

Since $\mathbb{N}$ is unbounded, we can find a value for $n$ that is bigger than $-\frac{1}{x}$. So this inequality is not true for every $1\leq n\mathbb{N}$. This means that if $x<0$ the element $x$ cannot belong to the intersection $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$.

Therefore we get that the intersection contains only the element $x=0$ and so we have $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )=\{0\}}$.

Is everything correct and complete?

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- We have that $(-n,n(\subseteq (-\infty, +\infty)$ for all $n\in \mathbb{N}$. So it follows that $\displaystyle{\bigcup_{n\in \mathbb{N}}\left (-n,n\right )\subseteq (-\infty, +\infty )}$.

Let $x\in (-\infty , +\infty)$. Now we have to prove that there is a $n$ such that $n>x$ and $x>-n$, or not? But how can we show that?

$$$$

- Let $\displaystyle{x\in \bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )}$. This means\begin{align*}&\forall n\in \mathbb{N} \ : \ x\in \left (n, 10n^2+50\right ) \\ & \Rightarrow \forall n\in \mathbb{N} \ : \ n < x< 10n^2+50 \end{align*}

We see that it must hold that $x> 0$. Now we want to solve for $n$.

From the first inequality we have $n<x$.

From the second inequality we have $x<10n^2+50 \Rightarrow 10n^2>x-50 \Rightarrow n^2>\frac{x}{10}-5$. From that we get $n>\sqrt{\frac{x}{10}-5}$, since $n$ is positiv and so the case $n<-\sqrt{\frac{x}{10}-5}$ is rejected, right?

So we have $$\forall n\in \mathbb{N} : \sqrt{\frac{x}{10}-5}<n<x$$ Since $\mathbb{N}$ is unbounded we can find always a value for $n$ that is bigger than $x$, or not?

Would that means that the intersection can't contain any element, and so $\displaystyle{\bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )=\emptyset}$ ?