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
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.4: Linear Transformations
Problem 24EQ
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F5de80a9c-52a0-4deb-b91c-ad55bd567dae%2F2q2f4c_processed.png&w=3840&q=75)
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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can you please re-explain "property of norm" step
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