Question

Let $K \subset C$. Prove that the following are equivalent: 1. K is closed and bounded. 2. Every open cover $K \subset \bigcup_{i \in I} U_i$ has a finite subcover $K \subset \bigcup_{i \in I_0} U_i$, for $I_0 \subset I$ finite. 3. Every sequence \{z_i\} $\subset K$ has a converging subsequence \{$z_{i_k}$\} (with \{$i_k$\} $\subset \mathbb{N}$) with limit in K. 4. For every continuous function $f: K \to \mathbb{R}$, the image $f(K)$ assumes its maximum.

          Let $K \subset C$. Prove that the following are equivalent:

1. K is closed and bounded.

2. Every open cover $K \subset \bigcup_{i \in I} U_i$ has a finite subcover $K \subset \bigcup_{i \in I_0} U_i$, for $I_0 \subset I$ finite.

3. Every sequence \{z_i\} $\subset K$ has a converging subsequence \{$z_{i_k}$\} (with \{$i_k$\} $\subset \mathbb{N}$) with limit in K.

4. For every continuous function $f: K \to \mathbb{R}$, the image $f(K)$ assumes its maximum.

Let K ⊂ C. Prove that the following are equivalent:

1. K is closed and bounded.

2. Every open cover K ⊂⋃i ∈ I Ui has a finite subcover K ⊂⋃i ∈ I0 Ui, for I0 ⊂ I finite.

3. Every sequence {zi} ⊂ K has a converging subsequence {zik} (with {ik} ⊂ℕ) with limit in K.

4. For every continuous function f: K →ℝ, the image f(K) assumes its maximum.

Added by Mitchell J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Shyam P

04/06/2022

Step 1

Assume K is closed and bounded. Let {Uᵢ} be an open cover of K. Since K is bounded, there exists a finite subcover {U₁, U₂, ..., Uₙ} that covers K. Therefore, K is covered by a finite subcover, which proves 2. Step 2: Next, we will prove that 2 implies 3. Show more…

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Let K be a subset of C. Prove that the following are equivalent: 1. K is closed and bounded. 2. Every open cover K ⊂ ⋃ᵢ∈I Uᵢ has a finite subcover K ⊂ ⋃ᵢ₌₁ᵤ Uᵢ, for I₀ ⊂ I finite. 3. Every sequence {zᵢ} ⊂ K has a converging subsequence {zᵢₖ} with {iₖ} ⊂ ℕ with limit in K. 4. For every continuous function f : K → ℝ, the image f(K) assumes its maximum.