Question

Prove that if \(f\) is continuous and \(f(a, b)<0,\) there exists a \(\delta\) -neighborhood about \((a, b)\) such that \(f(x, y)<0\) for every point \((x, y)\) in the neighborhood.

Short answer

For every \(\epsilon = -f(a, b) > 0\), there is a \(\delta > 0\) such that whenever \((x, y)\) is in the \(\delta\)-neighborhood of \((a, b)\), we have \(f(x, y) < 0\), thereby proving the necessitated property of the continuous function.

Step-by-step solution

Understanding the problem and establishing epsilon

The problem statement is connected to the concept of continuity which suggests that for every \(\epsilon > 0\), there is a \(\delta > 0\) such that \(|f(x) - f(a, b)| < \epsilon\) for all \((x, y)\) in the \(\delta\)-neighborhood about \((a, b)\). Here, since \(f(a, b) < 0\), let's take \(\epsilon = -f(a, b)\). This \(\epsilon\) is greater than 0 since \(f(a, b) < 0\).

Determine delta

By the definition of continuity, since \(f\) is continuous at \((a, b)\), there exists a \(\delta > 0\) such that for all \((x, y)\) in \(\delta\)-neighborhood about \((a, b)\), we have \(|f(x, y) - f(a, b)| < \epsilon\). This implies that \(-\epsilon < f(x, y) - f(a, b) < \epsilon\). Given that \(\epsilon = -f(a, b)\), the inequality becomes \(0 < f(x, y) < 2f(a, b)\).

Determine the function value in the neighborhood

As shown in the previous step, \(f(x, y) < 2f(a, b)\). Since \(f(a, b) < 0\), \(2f(a, b)\) is also less than 0. Therefore, we can deduce that \(f(x, y) < 0\) for every point \((x, y)\) in the \(\delta\)-neighborhood about \((a, b)\). Thus, the property is proven.