Request explain underlined portion 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 > 0such 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. SinceE₁ + E₂ = U{E₁ + 1₂ = 72 € E2},it follows that E₁ + E₂ is open.

It is given that X be a normed space, E1 is open in X and E2 be any subset of X. We have to explain…

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

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter4: Vector Spaces

Section4.CR: Review Exercises

Problem 73CR

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Request explain underlined portion

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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