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Chapter 2: Problem 15

Show that if \(a, b \in \mathbb{R}\), and \(a \neq b\), then there exist \(\varepsilon\) -neighborhoods \(U\) of \(a\) and \(V\) of \(b\) such that \(U \cap V=\emptyset\)

Short Answer

Expert verified
By defining \(\varepsilon=|a-b|/2\), and then defining the epsilon-neighborhoods U and V of a and b respectively as \(U = (a-\varepsilon, a+\varepsilon)\) and \(V = (b-\varepsilon, b+\varepsilon)\), it can be shown that a contradiction arises if there is assumed to be a common element in both U and V. Thus, U and V do not intersect (i.e., \(U \cap V = \emptyset\)).

Step by step solution

01

Define epsilon

Firstly, since \(a \neq b\), we can note that the value |a - b| must be positive. Let's define \(\varepsilon=|a-b|/2\). Given that the modulus of a real number is always positive, this ensures that \(\varepsilon > 0\).
02

Define U and V

We now define the neighborhoods U and V around a and b respectively. Define \(U = (a-\varepsilon, a+\varepsilon)\) and \(V = (b-\varepsilon, b+\varepsilon)\).
03

Show U and V have no common elements

Assume that there is a point that lies in both U and V. Then this point must satisfy both \(a-\varepsilon < x < a+\varepsilon\) and \(b-\varepsilon < x < b+\varepsilon\). This can only happen if \(\varepsilon > |a-b|/2\). This contradicts our definition of \(\varepsilon=|a-b|/2\). Therefore, U and V cannot have common elements, and so \(U \cap V = \emptyset\).

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