Question

Consider the following topology on $\mathbf{B}$. To define a neighborhood of $A$ in $\mathbf{B}$, fix $n$ vectors, $h_1, \ldots, h_n$, and positive $\epsilon$. Consider all $B$ in $\mathbf{B}$ with $\left\|(A-B)\left(h_1\right)\right\| \leq \epsilon, \ldots,\left\|(A-B)\left(h_n\right)\right\| \leq \epsilon$. Prove that this defines a topology on B and that theorem 63 refers to convergence in this topology.

   Consider the following topology on $\mathbf{B}$. To define a neighborhood of $A$ in $\mathbf{B}$, fix $n$ vectors, $h_1, \ldots, h_n$, and positive $\epsilon$. Consider all $B$ in $\mathbf{B}$ with $\left\|(A-B)\left(h_1\right)\right\| \leq \epsilon, \ldots,\left\|(A-B)\left(h_n\right)\right\| \leq \epsilon$. Prove that this defines a topology on B and that theorem 63 refers to convergence in this topology.

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Mathematical physics

Mathematical physics

Robert Geroch 1st Edition

Chapter 53, Problem 366

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Step 1

We are given a set $\mathbf{B}$ and a way to define neighborhoods around elements $A \in \mathbf{B}$. The neighborhoods are defined using $n$ vectors $h_1, \ldots, h_n$ and a positive $\epsilon$. For a given $A \in \mathbf{B}$, the neighborhood consists of all $B Show more…

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Consider the following topology on $\mathbf{B}$. To define a neighborhood of $A$ in $\mathbf{B}$, fix $n$ vectors, $h_1, \ldots, h_n$, and positive $\epsilon$. Consider all $B$ in $\mathbf{B}$ with $\left\|(A-B)\left(h_1\right)\right\| \leq \epsilon, \ldots,\left\|(A-B)\left(h_n\right)\right\| \leq \epsilon$. Prove that this defines a topology on B and that theorem 63 refers to convergence in this topology.

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