Denseness And The Limit Of A Sequence - An Example
Imagine you are a certain distance from an objective. Lets say this distance is 16 units. You are allowed to make progress toward your objective by cutting the distance in half at any time, as many times as you like. When will you reach your objective? In order to answer this question we must evaluate a limit. But before doing that, we may ask ourselves another question: how many times can we divide the distance?
The distance in this case, at each step, is a rational number. A rational number is a quotient of two whole numbers in the form n/m (hence the number system is denoted “Q”). So our problem is within Q, the set of rational numbers. Mathematically we have distance $d=\frac{16}{2^n}$ where n is our step number. At one step $d=8$, at 2 steps $d=4$, and then $2$,$1$,$\frac{1}{2}$,$\frac{1}{4}$, etc.
Our progress, one step at the time, is called a sequence (in Q). A sequence means we can denumerate the step number $n$. We can assign an integer number to each step. It turns out that because Q is dense, we can divide the number infinitely many times. So dense means you can pick any two numbers and always find yet another number in between. This concept is fundamentally important. It means the space Q is infinite. It also means that any subdivision, any interval of Q which we define as the set of points in Q between two numbers that are not the same, no matter how small, is still infinite. Despite this denseness, Q is not complete but we'll get to that later. Going back to our distance, it is clear that it gets closer and closer to zero as we increase the step number but it never quite gets there for any finite number $n$ of steps. This is where the limit comes into play. In this case, the limit is the number the sequence comes closest to as the number of steps moves to infinity. The distance here converges to zero at infinite steps. Per definition, a sequence converges if it comes closer and closer to a single point as we increase the steps.
The objective is reached in the limit of infinite steps.
Denseness, completeness, sequences and limits are all fundamental concepts in analysis.
The distance in this case, at each step, is a rational number. A rational number is a quotient of two whole numbers in the form n/m (hence the number system is denoted “Q”). So our problem is within Q, the set of rational numbers. Mathematically we have distance $d=\frac{16}{2^n}$ where n is our step number. At one step $d=8$, at 2 steps $d=4$, and then $2$,$1$,$\frac{1}{2}$,$\frac{1}{4}$, etc.
Our progress, one step at the time, is called a sequence (in Q). A sequence means we can denumerate the step number $n$. We can assign an integer number to each step. It turns out that because Q is dense, we can divide the number infinitely many times. So dense means you can pick any two numbers and always find yet another number in between. This concept is fundamentally important. It means the space Q is infinite. It also means that any subdivision, any interval of Q which we define as the set of points in Q between two numbers that are not the same, no matter how small, is still infinite. Despite this denseness, Q is not complete but we'll get to that later. Going back to our distance, it is clear that it gets closer and closer to zero as we increase the step number but it never quite gets there for any finite number $n$ of steps. This is where the limit comes into play. In this case, the limit is the number the sequence comes closest to as the number of steps moves to infinity. The distance here converges to zero at infinite steps. Per definition, a sequence converges if it comes closer and closer to a single point as we increase the steps.
The objective is reached in the limit of infinite steps.
Denseness, completeness, sequences and limits are all fundamental concepts in analysis.
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