Cardinality of an interval as a limit

[Fermat1]

Let $I$ be an interval and $A_{n}$ be the set of $k/n$ where $k$ is an integer.

Prove that $|I|$ is the limit as $n$ tends to infinity of $\frac{1}{n}|(IA_{n})|$ where $IA_{n}$ denotes intersection.

My plan was to split it up into cases for the different type of intervals and come up with formulas for $|(IA_{n})|$, but I'm finding that very tricky.

Thanks

 

[Euge]

Let $I$ be an interval and $A_{n}$ be the set of $k/n$ where $k$ is an integer.

Prove that $|I|$ is the limit as $n$ tends to infinity of $\frac{1}{n}|(IA_{n})|$ where $IA_{n}$ denotes intersection.

My plan was to split it up into cases for the different type of intervals and come up with formulas for $|(IA_{n})|$, but I'm finding that very tricky.

Thanks

Hi Fermat,

The stated problem is a bit unclear; you used the absolute value bars to indicate both cardinality and Lebesgue measure? I'm assuming that $|I|$ is the Lebesgue measure of $I$ and $|IA_n|$ is the cardinality of $I\cap A_n$ (with the exception that |IA_n| is $\infty$ if $I\cap A_n$ is infinte). If $I$ is an unbounded interval, $|I| = \infty$ and $|I\cap A_n| = \infty$. So the equation will hold. So suppose $I$ is a bounded interval. There are four possible forms for $I$: $(a,b)$, $[a,b)$, $(a,b]$, and $[a,b]$. I'll do the case when $I = [a,b]$, as the others have similar arguments. When $I = [a,b]$, $|I \cap A_n|$ is no more than $nb - na + 1$ and no less than $nb - na - 3$. Hence, $||I \cap A_n| - (nb - na)| < C$ for some constant $C$ independent of $n$. Thus,

\(\displaystyle \left|\frac{|IA_n|}{n} - |I|\right| < \frac{C}{n}.\)

Since $\frac{C}{n} \to 0$ as $n \to \infty$, we deduce that

\(\displaystyle \lim_{n\to \infty} \frac{|IA_n|}{n} = |I|\).

 

[Fermat1]

Hi Fermat,

The stated problem is a bit unclear; you used the absolute value bars to indicate both cardinality and Lebesgue measure? I'm assuming that $|I|$ is the Lebesgue measure of $I$ and $|IA_n|$ is the cardinality of $I\cap A_n$. If $I$ is an unbounded interval, $|I| = \infty$ and $|I\cap A_n| = \infty$. So the equation will hold. So suppose $I$ is a bounded interval. There are four possible forms for $I$: $(a,b)$, $[a,b)$, $(a,b]$, and $[a,b]$. I'll do the case when $I = [a,b]$, as the others have similar arguments. When $I = [a,b]$, $I \cap A_n$ has no more than $nb - na + 1$ elements and no less than $nb - na - 3$ elements. Hence, $||I \cap A_n| - (nb - na)| < C$ for some constant $C$ independent of $n$. Thus,

\(\displaystyle \left|\frac{|IA_n|}{n} - |I|\right| < \frac{C}{n}.\)

Since $\frac{C}{n} \to 0$ as $n \to \infty$, we deduce that

\(\displaystyle \lim_{n\to \infty} \frac{|IA_n|}{n} = |I|\).

Thanks very much. Can I ask how you got the bounds $nb - na + 1$ and $nb - na - 3$, and how I obtain bounds for the other sorts of intervals. Thanks

 

[Euge]

Thanks very much. Can I ask how you got the bounds $nb - na + 1$ and $nb - na - 3$, and how I obtain bounds for the other sorts of intervals. Thanks

When $I = [a, b]$, $I\cap A_n$ is the set of rationals of the form $\frac{k}{n}$ which satisfy $a \le \frac{k}{n} \le b$, i.e., $na \le k \le nb$. So $|I \cap A_n|$ is the number of integers $k$ in the interval $[na,nb]$. The least integer in $[na,nb]$ is either $na$ or $\lfloor na \rfloor + 1$ (depending on whether or not $na$ is an integer) and similarly the greatest integer in $[na,nb]$ is either $nb$ or $\lfloor nb \rfloor - 1$. Hence $|I \cap A_n|$ is either $nb - na + 1$, $\lfloor nb \rfloor - na$, $nb - \lfloor na \rfloor$, or $\lfloor nb \rfloor - \lfloor na \rfloor - 2$. Each of these numbers are between $nb - na - 3$ and $nb - na + 1$ since $\lfloor nb \rfloor \in (nb - 1, nb]$ and $\lfloor na \rfloor \in (na - 1, na]$. Therefore $|I \cap A_n|$ is no more than $nb - na + 1$ than $nb - na + 3$.

The other cases are dealt with in nearly the same way.

 

[blue_raver22]

for your question! I understand your approach of breaking down the problem into different cases for different types of intervals. However, there is a simpler and more general approach to proving this statement.

First, let's define the cardinality of a set $S$ as the number of elements in that set, denoted by $|S|$. In this case, we are interested in the cardinality of an interval $I$, which is the number of real numbers contained within that interval.

Next, let's define the set $IA_{n}$ as the intersection of the interval $I$ and the set $A_{n}$. In other words, $IA_{n}$ is the set of all real numbers that are both in the interval $I$ and in the set $A_{n}$. We can write this as $IA_{n} = \{x \in I | x = \frac{k}{n}, k \in \mathbb{Z}\}$.

Now, as $n$ tends to infinity, the set $A_{n}$ becomes denser and denser, meaning that there are more and more real numbers in $A_{n}$ that can be expressed as $\frac{k}{n}$ for some integer $k$. As a result, the intersection $IA_{n}$ will also become denser and closer to the interval $I$ as $n$ tends to infinity.

This leads us to the key idea that the limit of $\frac{1}{n}|(IA_{n})|$ as $n$ tends to infinity is equal to the cardinality of the interval $I$, or $|I|$. This is because as $n$ tends to infinity, the ratio of $|(IA_{n})|$ to $n$ will approach the cardinality of $I$, which is the number of real numbers in $I$.

To prove this statement, we can use the concept of limits. By definition, the limit of a sequence $a_{n}$ is a number $L$ such that for any small positive number $\epsilon$, there exists a positive integer $N$ such that for all $n > N$, the value of $a_{n}$ will be within $\epsilon$ distance from $L$. In other words, as $n$ gets larger and larger, the value of $a_{n}$ gets closer and closer to $L$.

In our case, as $