Sure, let's go through this step by step.

1. Show that any bounded subset of R is totally bounded:

A set S in R is bounded if there exists a real number M such that |x| ≤ M for all x in S. Now, let's show that S is totally bounded. By definition, a set is totally bounded if for every ε 0, there exists a finite number of balls of radius ε that can cover the set.

Since S is bounded, there exists a ball B(M) of radius M that can cover S. Now, for any ε 0, we can always find a finite number of balls of radius ε that can cover B(M), and hence cover S. This is because the real line R is Archimedean, meaning for any real numbers x and y with x 0, there exists a natural number n such that nx y. In this case, we can always find a natural number n such that nε 2M, meaning we can cover B(M) with n balls of radius ε. Therefore, any bounded subset of R is totally bounded.

2. Is this true in R^n?

Yes, this is also true in R^n. The proof is similar. A set S in R^n is bounded if there exists a real number M such that ||x|| ≤ M for all x in S, where ||x|| denotes the Euclidean norm. For any ε 0, we can always find a finite number of balls of radius ε that can cover the ball B(M) of radius M, and hence cover S. This is because R

Question

Sure, let's go through this step by step.

1. Show that any bounded subset of R is totally bounded:

A set S in R is bounded if there exists a real number M such that |x| ≤ M for all x in S. Now, let's show that S is totally bounded. By definition, a set is totally bounded if for every ε > 0, there exists a finite number of balls of radius ε that can cover the set.

Since S is bounded, there exists a ball B(M) of radius M that can cover S. Now, for any ε > 0, we can always find a finite number of balls of radius ε that can cover B(M), and hence cover S. This is because the real line R is Archimedean, meaning for any real numbers x and y with x > 0, there exists a natural number n such that nx > y. In this case, we can always find a natural number n such that nε > 2M, meaning we can cover B(M) with n balls of radius ε. Therefore, any bounded subset of R is totally bounded.

2. Is this true in R^n?

Yes, this is also true in R^n. The proof is similar. A set S in R^n is bounded if there exists a real number M such that ||x|| ≤ M for all x in S, where ||x|| denotes the Euclidean norm. For any ε > 0, we can always find a finite number of balls of radius ε that can cover the ball B(M) of radius M, and hence cover S. This is because R

Knowee AI · Accepted Answer

Sure, let's go through this step by step.

1. Show that any bounded subset of R is totally bounded:

A set S in R is bounded if there exists a real number M such that |x| ≤ M for all x in S. Now, let's show that S is totally bounded. By definition, a set is totally bounded if for every ε > 0, there exists a finite number of balls of radius ε that can cover the set.

Since S is bounded, there exists a ball B(M) of radius M that can cover S. Now, for any ε > 0, we can always find a finite number of balls of radius ε that can cover B(M), and hence cover S. This is because the real line R is Archimedean, meaning for any real numbers x and y with x > 0, there exists a natural number n such that nx > y. In this case, we can always find a natural number n such that nε > 2M, meaning we can cover B(M) with n balls of radius ε. Therefore, any bounded subset of R is totally bounded.

2. Is this true in R^n?

Yes, this is also true in R^n. The proof is similar. A set S in R^n is bounded if there exists a real number M such that ||x|| ≤ M for all x in S, where ||x|| denotes the Euclidean norm. For any ε > 0, we can always find a finite number of balls of radius ε that can cover the ball B(M) of radius M, and hence cover S. This is because R

show that any bounded subset of R is totally bounded . is this true in R^n? can you make the same statement in any metric space?

Question

Show that any bounded subset of $\mathbb{R}$ is totally bounded. Is this true in $\mathbb{R}^n$ ? Can you make the same statement in any metric space?

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