Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function and let \( J \) be a bounded open interval in \( \mathbb{R} \). Define 
\[ W(f, J) = \sup \{ f(x) | x \in J \} - \inf \{ f(x) | x \in J \} \] Which one of the following is FALSE? 
 

• A. \( W(f, J_1) \leq W(f, J_2) \) if \( J_1 \subset J_2 \)
• B. If \( f \) is a bounded function in \( J \) and \( J_1 \supset J_2 \supset \cdots \supset J_n \supset \cdots\) such that the length of the interval \( J_n \) tends to 0 as \( n \to \infty \), then \( \lim_{n \to \infty} W(f, J_n) = 0 \)
• C. If \( f \) is discontinuous at a point \( a \in J \), then \( W(f, J) \neq 0 \)
• D. If \( f \) is continuous at a point \( a \in J \), then for any given \( \epsilon > 0 \) there exists an interval \( I \subset J \) such that \( W(f, I) < \epsilon \)

Hint:

When working with functions, the supremum and infimum are influenced by the continuity of the function. Discontinuities do not always imply a non-zero difference between the supremum and infimum.

Answer 1

The Correct Option is B

Given: $W(f,J) = \sup{f(x) \mid x \in J} - \inf{f(x) \mid x \in J}$

This is the oscillation or width of function $f$ on interval $J$.

Option (A): $W(f,J_1) \leq W(f,J_2)$ if $J_1 \subset J_2$

If $J_1 \subset J_2$, then:

$\sup{f(x) \mid x \in J_1} \leq \sup{f(x) \mid x \in J_2}$

$\inf{f(x) \mid x \in J_1} \geq \inf{f(x) \mid x \in J_2}$

Therefore: $$W(f,J_1) = \sup_{J_1} f - \inf_{J_1} f \leq \sup_{J_2} f - \inf_{J_2} f = W(f,J_2)$$

Option (A) is TRUE 

Option (B): If $f$ is bounded in $J$ and $J \supset J_1 \supset J_2 \supset \cdots \supset J_n \supset \cdots$ with length tending to 0, then $\lim_{n \to \infty} W(f,J_n) = 0$

This statement claims that for any nested sequence of intervals whose lengths shrink to 0, the oscillation must approach 0.

Counterexample:

Consider the function: $$f(x) = \begin{cases} 1, & x \text{ is rational} \ 0, & x \text{ is irrational} \end{cases}$$

This is the Dirichlet function, which is bounded.

For any interval $J_n$ (no matter how small), we have:

$\sup{f(x) \mid x \in J_n} = 1$ (since every interval contains rationals)

$\inf{f(x) \mid x \in J_n} = 0$ (since every interval contains irrationals)

Therefore: $W(f,J_n) = 1 - 0 = 1$ for all $n$

So $\lim_{n \to \infty} W(f,J_n) = 1 \neq 0$

Option (B) is FALSE 

Option (C): If $f$ is discontinuous at a point $a \in J$, then $W(f,J) \neq 0$

If $f$ is discontinuous at $a \in J$, then there exists $\epsilon > 0$ such that for every $\delta > 0$, there exist points $x, y$ within $\delta$ of $a$ where $|f(x) - f(y)| \geq \epsilon$.

Since $a \in J$ and $J$ is open, we can find such points in $J$, which means: $$W(f,J) \geq |f(x) - f(y)| \geq \epsilon > 0$$

Therefore $W(f,J) \neq 0$.

Option (C) is TRUE 

Option (D): If $f$ is continuous at $a \in J$, then for any $\epsilon > 0$ there exists an interval $I \subset J$ such that $W(f,I) < \epsilon$

If $f$ is continuous at $a$, then for any $\epsilon > 0$, there exists $\delta > 0$ such that: $$|x - a| < \delta \implies |f(x) - f(a)| < \frac{\epsilon}{2}$$

Choose $I = (a - \delta/2, a + \delta/2) \cap J$. For any $x, y \in I$: $$|f(x) - f(y)| \leq |f(x) - f(a)| + |f(a) - f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Therefore: $$W(f,I) = \sup_{x \in I} f(x) - \inf_{x \in I} f(x) < \epsilon$$

Option (D) is TRUE 

Answer: (B)