Problem 1 Show that a convergent sequence ... [FREE SOLUTION]

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Elementary Real Analysis

Brian S. Thomson, Judith B. Bruckner, Andrew M. Bruckner

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 13: Problem 1

Show that a convergent sequence in a metric space must be a Cauchy sequence.

Short Answer

Expert verified

A convergent sequence is a Cauchy sequence because distances can be bounded using the triangle inequality and convergence properties.

Step by step solution

Understand the Definitions

A sequence $ (x_n) $ in a metric space $(X, d)$ is said to be convergent if there exists a point $x \in X$ such that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(x_n, x) < \epsilon$ for all $n \geq N$. A sequence is a Cauchy sequence if for every $\epsilon > 0$, there exists an $M \in \mathbb{N}$ such that $d(x_m, x_n) < \epsilon$ for all $m, n \geq M$.

Use Convergence to Find N

Since the sequence $ (x_n) $ converges to $ x $, for an arbitrary $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(x_n, x) < \epsilon/2$ for all $n \geq N$. This is by the definition of convergence.

Apply the Triangle Inequality

For $m, n \geq N$, use the triangle inequality to bind the distance between $x_m$ and $x_n$: $d(x_m, x_n) \leq d(x_m, x) + d(x_n, x)$.

Substitute with Convergence Condition

From the convergence condition, we know $d(x_m, x) < \epsilon/2$ and $d(x_n, x) < \epsilon/2$ for $m, n \geq N$. Substitute these into the inequality to get: $d(x_m, x_n) \leq \epsilon/2 + \epsilon/2 = \epsilon$.

Conclude Cauchy Condition

Since $d(x_m, x_n) < \epsilon$ for all $m, n \geq N$, the sequence $ (x_n) $ satisfies the Cauchy condition, meaning it is a Cauchy sequence by definition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Sequence

In a metric space, a sequence $(x_n)$is said to converge to a point $x$ if, as you progress further along the sequence, the terms get closer and closer to $x$. Intuitively, imagine walking towards a specific point along a path. The further you walk, the closer you get to that point.
For a sequence to be convergent, for every tiny positive distance you can think of, there exists a position in the sequence beyond which all terms are closer to $x$ than that distance.
This shows stability, as beyond a certain point $N$, the sequence almost stays within an $\epsilon$ range of $x$, indicating it's heading towards a predictable target. This behavior helps to understand and predict patterns in mathematical sequences.

Cauchy Sequence

A Cauchy sequence is a sequence wherein terms get arbitrarily close to each other beyond a certain point. Think about friends meeting at a landmark. As they get closer, the distance between any two of them becomes small and, ideally, they converge to meet up.
More formally, for any small distance $\epsilon > 0$, you can find a position $M$ such that any part of the sequence beyond $M$ has terms within distance $\epsilon$ of each other. This means the sequence becomes tightly packed as it progresses.

Cauchy sequences help to ensure the existence of a limit, even if it is not explicitly known.
In metric spaces, every convergent sequence is Cauchy, since they both get closer around a specific point, making the terms get closer to each other.

Triangle Inequality

One of the essential properties in a metric space is the triangle inequality. It helps us understand how to measure distances indirectly by involving a third point. Think of it like taking two sides of a triangle: the sum is always greater than or equal to the remaining side.
Mathematically, for any three points $a, b,$ and $c$, the inequality states:\[ d(a, c) \leq d(a, b) + d(b, c) \]
In the convergent sequence proof, triangle inequality is instrumental. It gives the bound needed to show that the terms in a sequence become very close to each other, confirming its Cauchy nature.

This inequality allows distance relations to be flexible, useful to derive properties of convergence and compactness in a space.
It's a cornerstone that simplifies complex relations into manageable bounds in metrics.

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91影视

Show that a convergent sequence in a metric space must be a Cauchy sequence.

Short Answer

Step by step solution

Understand the Definitions

Use Convergence to Find N

Apply the Triangle Inequality

Substitute with Convergence Condition

Conclude Cauchy Condition

Key Concepts

Convergent Sequence

Cauchy Sequence

Triangle Inequality

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