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Let X be a normed space. (a) If E₁ is open in X and E₂ C X, then E₁ + E₂ is open in X. Proof: (a) Let x € X and r₁ € E₁. Since E₁ is open, there is some r > 0 such that U(₁,7) CE₁. But U(x₁+x,r) = U(1₁, r) + x C E₁ +1. Hence E₁ + is open for every z EX. Since

Let X be a normed space. (a) If E₁ is open in X and E₂ C X, then E₁ + E₂ is open in X. Proof: (a) Let x € X and r₁ € E₁. Since E₁ is open, there is some r > 0 such that U(₁,7) CE₁. But U(x₁+x,r) = U(1₁, r) + x C E₁ +1. Hence E₁ + is open for every z EX. Since

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let X be a normed space. (a) If E₁ is open in X and E₂ C X, then E₁ + E2 is open in X. Proof: (a) Let re X and r₁ € E₁. Since E1₁ is open, there is some r > 0 such that U(₁,7) CE₁. But U(x₁ + x,r) = U(x₁, r) + x C E₁ + x. Hence E₁ + is open for every x E X. Since E₁ + E₂ = U{E₁ + 1₂ = 72 € E2}, it follows that E₁ + E₂ is open.
Transcribed Image Text:Let X be a normed space. (a) If E₁ is open in X and E₂ C X, then E₁ + E2 is open in X. Proof: (a) Let re X and r₁ € E₁. Since E1₁ is open, there is some r > 0 such that U(₁,7) CE₁. But U(x₁ + x,r) = U(x₁, r) + x C E₁ + x. Hence E₁ + is open for every x E X. Since E₁ + E₂ = U{E₁ + 1₂ = 72 € E2}, it follows that E₁ + E₂ is open.
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