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[SOLVED] Intesection of intervals

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,114
Hey!! :eek:

I want to determine the following sets:
  1. $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$
  2. $\displaystyle{\bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$
  3. $\displaystyle{\bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )}$



I have done the following:
  1. 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? (Wondering)

    $$$$
  2. 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? (Wondering)

    $$$$
  3. 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 $\displaystyle{\bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )=\emptyset}$ ? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
1. $\displaystyle{\bigcap_{1\leq n\in \mathbb{N}}\left (-\frac{1}{n},\frac{1}{n}\right )}$

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?
Looks fine to me. (Nod)

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

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?
Can we pick $n = \lceil \operatorname{abs}(x) \rceil + 1$? (Wondering)


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

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 $\displaystyle{\bigcap_{n\in \mathbb{N}}\left (n, 10n^2+50\right )=\emptyset}$ ?
Yep.

I think we can do it a bit quicker.
The intervals have a lower bound of $n$.
If some $x$ would be in the intersection, then $x$ must be in all intervals. But $x$ will not be in any interval where $n>x$, so it won't be in the intersection. This is a contradiction, so the intersection must be the empty set. (Nerd)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,114
But $x$ will not be in any interval where $n>x$, so it won't be in the intersection.
Could you explain that to me further?

Do we consider two intervals and then we show that $x$ cannot belong to both?

(Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Could you explain that to me further?

Do we consider two intervals and then we show that $x$ cannot belong to both?
The intersection is of the form:
$$(1,a_1)\cap(2,a_2)\cap(3,a_3)\cap\ldots\cap(n,a_n)\cap\ldots$$
For an $x$ to be in this intersection, it must be in all intervals.
But there will always be an interval that starts with an $n$ that is higher than $x$ is. (Thinking)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,114
The intersection is of the form:
$$(1,a_1)\cap(2,a_2)\cap(3,a_3)\cup\ldots\cup(n,a_n)\cup\ldots$$
For an $x$ to be in this intersection, it must be in all intervals.
But there will always be an interval that starts with an $n$ that is higher than $x$ is. (Thinking)
Ahh I got it now!! (Malthe)


As for 2. if we pick $n = \lceil \operatorname{abs}(x) \rceil + 1$ then we have that $x\in (-n,n)$. But how do we get the union? I got stuck right now. (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
As for 2. if we pick $n = \lceil \operatorname{abs}(x) \rceil + 1$ then we have that $x\in (-n,n)$. But how do we get the union? I got stuck right now.
For an $x$ to be in the union, it must be in at least one of the intervals.
And we just found an interval that it is in, didn't we? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,114
For an $x$ to be in the union, it must be in at least one of the intervals.
And we just found an interval that it is in, didn't we? (Wondering)
Ahh yes!!

So we have the following:

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 $x\in (-\infty , +\infty)$. For $n = \lceil \operatorname{abs}(x) \rceil + 1$ it holds that $x\in (-n,n)$. Therefore it holds that $\displaystyle{x\in \bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$ and so we get $\displaystyle{(-\infty, +\infty )\subseteq \bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$.

Therefore $\displaystyle{\bigcup_{n\in \mathbb{N}}\left (-n,n\right )= (-\infty, +\infty )}$.


Is everything correct? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Ahh yes!!

So we have the following:

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 $x\in (-\infty , +\infty)$. For $n = \lceil \operatorname{abs}(x) \rceil + 1$ it holds that $x\in (-n,n)$. Therefore it holds that $\displaystyle{x\in \bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$ and so we get $\displaystyle{(-\infty, +\infty )\subseteq \bigcup_{n\in \mathbb{N}}\left (-n,n\right )}$.

Therefore $\displaystyle{\bigcup_{n\in \mathbb{N}}\left (-n,n\right )= (-\infty, +\infty )}$.


Is everything correct?
Yep. (Nod)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,114