Question

Prove: If \(f\) assumes only finitely many values, then \(f\) is continuous at a point \(x_{0}\) in \(D_{f}^{0}\) if and only if \(f\) is constant on some interval \(\left(x_{0}-\delta, x_{0}+\delta\right) .\)

Short answer

Question: Prove that a function \(f\) assumes only finitely many values and is continuous at a point \(x_{0}\) in its domain if and only if it is constant on some interval \((x_{0}-\delta, x_{0}+\delta)\). Answer: The proof consists of two parts: 1. If \(f\) is continuous at \(x_{0}\), then it is constant on some interval \((x_{0}-\delta, x_{0}+\delta)\). This is shown by considering the maximum difference between any two values of \(f\) and applying the definition of continuity. 2. If \(f\) is constant on some interval \((x_{0}-\delta, x_{0}+\delta)\), then it is continuous at \(x_{0}\). This follows from the fact that \(f(x) = f(x_{0})\) for all \(x\) in the interval, and thus the definition of continuity is satisfied. Overall, the proof shows that a function which assumes only finitely many values and is continuous at a point must be constant on some interval around that point.

Step-by-step solution

Prove that if \(f\) is continuous at \(x_{0}\), then it is constant on some interval \((x_{0}-\delta, x_{0}+\delta)\)

Assume that \(f\) is continuous at \(x_{0}\). By definition of continuity, for any given \(\epsilon > 0\), there exists a \(\delta > 0\) such that $$ \left| f(x) - f(x_{0}) \right| < \epsilon $$ whenever \(0 < \left| x - x_{0} \right| < \delta\), with \(x \in D_f^0\). Since \(f\) assumes only finitely many values, let \(M = \max{\left\{ \left| f(x)-f(y) \right|.\ x,y \in D_f^0 \right\}}\). Consider \(\epsilon = \frac{M}{2}\). There exists a \(\delta > 0\) such that, $$ \left| f(x) - f(x_{0}) \right| < \frac{M}{2} $$ whenever \(0 < \left| x - x_{0} \right| < \delta\). But \(M\) is the maximum difference between any two values of \(f\). Therefore, \(\left| f(x) - f(x_{0}) \right| < \frac{M}{2}\) implies \(f(x) = f(x_{0})\) for all \(x \in (x_{0} - \delta, x_{0} + \delta)\). Thus, \(f\) is constant on the interval \((x_{0} - \delta, x_{0} + \delta)\).

Prove that if \(f\) is constant on some interval \((x_{0}-\delta, x_{0}+\delta)\), then it is continuous at \(x_{0}\)

Assume that \(f\) is constant on the interval \((x_{0}-\delta, x_{0}+\delta)\). We need to show that \(f\) is continuous at \(x_{0}\). We can do this by checking the definition of continuity. For any given \(\epsilon > 0\), we have to find a \(\delta > 0\) such that $$ \left| f(x) - f(x_{0}) \right| < \epsilon $$ whenever \(0 < \left| x - x_{0} \right| < \delta\), with \(x \in D_f^0\). Since \(f\) is constant on the interval \((x_{0} - \delta, x_{0} + \delta)\), we know that \(f(x) = f(x_{0})\) for all \(x \in (x_{0} - \delta, x_{0} + \delta)\). Therefore, $$ \left| f(x) - f(x_{0}) \right| = 0 < \epsilon $$ for all \(x \in (x_{0} - \delta, x_{0} + \delta)\). This implies that \(f\) is continuous at \(x_{0}\). Hence, we have shown both directions of the equivalence, which completes the proof.

Key concepts

Continuous Functions

When we dive into real analysis, a pivotal concept we encounter is that of continuous functions. These are functions that, intuitively, have no abrupt 'jumps' or 'gaps' in their graphs, and they flow smoothly when plotted. In more technical terms, a function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at the point.

Continuity can be visually understood when drawing the function without lifting the pencil from the paper. For students solving exercises, it’s important to recognize that continuous functions preserve the closeness of points: values at nearby points on the domain remain close to each other on the range. This intuitive understanding helps when learning the more formal, rigorous definitions of continuity, such as the epsilon-delta definition.

Epsilon-Delta Definition of Continuity

The epsilon-delta definition of continuity is the formal, mathematical way to define a function being continuous at a point that doesn't rely on graphs or visual aids. Here's how it works: a function f is continuous at a point x_0 if for every small number \(\bsp\epsilon  \gt; 0\), there’s another small number \(\ \delta\ \ \> 0\) so that whenever the distance between x and x_0 is less than \(\delta\), the difference in their outputs f(x) and f(x_0) is less than \(\epsilon\).

In simpler terms, no matter how close we want the function values to be near x_0 (that's our \(\epsilon\)), we can always find an interval around x_0 so that the values of f within this interval remain within the desired proximity to f(x_0). This concept assures us that the function behaves 'nicely' and predictably around x_0, cementing the concept of continuity.

Constant Function Property

Let’s zoom into a specific scenario with the constant function property. A constant function is a function that returns the same value no matter what input it is given within the domain. This implies a sort of 'ultimate' continuity — if the function doesn't change at all, it certainly won't leap or have breaks.

In the problem at hand, we encountered that if a function assumes only finitely many values, continuity at a point implies it must be constant over some interval around that point. This stems from the fact that there's a limit to how 'different' the function can be around that point—if it were to change, given the finite set of values, it would eventually 'jump'. This is neatly tied to the notion that within a very tiny region, the function can't do anything strange; it’s stuck being constant.

Investigating such properties enhances students' understanding of the different layers of continuity and solidifies the relationship between the abstract definition and practical implications in the behaviors of functions.