Question Statement

Prove that $f(x)=\{\begin{array}{ll}1,& x\text{ rational }\\ 0,& x\text{ irrational }\end{array}$ is continuous at every real number. What does the graph of $f$ looks like?

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Background and Explanation

This problem concerns the Dirichlet function, a fundamental counterexample in real analysis. The key prerequisite is understanding that both rational and irrational numbers are dense in $\mathbb{R}$—meaning every open interval, no matter how small, contains infinitely many numbers of each type.

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Solution

Part 1: Analysis of Continuity

Given the function: $f(x)=\{\begin{array}{ll}1,& x\text{ rational }\\ 0,& x\text{ irrational }\end{array}$

Let $x=a$ be any arbitrary real number.

If $a$ is rational, then $f(a)=1$. If $a$ is irrational, then $f(a)=0$.

When we take the limit as $x$ approaches $a$, we encounter both rational and irrational numbers in any neighbourhood of $a$ (due to the density of $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ in $\mathbb{R}$).

This means that as $x$ approaches $a$, the function values oscillate between $1$ (at rational points) and $0$ (at irrational points). Therefore: ${\lim }_{x\to a}f(x)\text{ does not exist}$

Since the limit fails to exist at $x=a$, and $x=a$ was chosen as an arbitrary real number, we conclude that $f(x)$ is discontinuous for all $x\in \mathbb{R}$.

(Note: The function is actually discontinuous everywhere, contrary to the original question's premise.)

Part 2: Description of the Graph

The graph of $f(x)$ consists of two horizontal lines:

• Since $f(x)=1$ when $x$ is rational, one line is $y=1$, representing the function's value for all rational numbers. This appears as a dense collection of uncountably many distinct points along the line $y=1$.

• Since $f(x)=0$ when $x$ is irrational, the other line is $y=0$, representing the function's value for all irrational numbers. This similarly appears as a dense collection of points along $y=0$.

Due to the density of both rational and irrational numbers on the real number line, these two lines appear as dense collections of points with no gaps, but there is no connection between the two levels. It is not possible to draw a continuous curve that represents this function.

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Key Formulas or Methods Used

• Definition of continuity at a point: ${\lim }_{x\to a}f(x)=f(a)$
• Density property of real numbers: Every open interval $(a-\delta ,a+\delta )$ contains both rational and irrational numbers for any $\delta >0$
• Criterion for limit existence: A limit ${\lim }_{x\to a}f(x)$ exists only if $f(x)$ approaches a single unique value as $x\to a$
• Dirichlet function: The characteristic function ${\chi }_{\mathbb{Q}}(x)=\{\begin{array}{ll}1& x\in \mathbb{Q}\\ 0& x\notin \mathbb{Q}\end{array}$

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Summary of Steps

1. State the function definition: Identify $f(x)$ as the indicator function for rational numbers.
2. Select an arbitrary point: Let $x=a$ represent any real number (rational or irrational).
3. Determine $f(a)$: Note that $f(a)=1$ if $a\in \mathbb{Q}$ and $f(a)=0$ if $a\notin \mathbb{Q}$.
4. Examine the limit process: Recognize that any neighborhood of $a$ contains both rational and irrational preimages.
5. Conclude non-existence of limit: Since $f(x)$ takes both values $0$ and $1$ arbitrarily close to $a$, the limit cannot exist.
6. Establish discontinuity: Conclude $f$ is discontinuous at every real number since the limit never equals the function value.
7. Describe the graph: Identify the two dense sets of points at $y=1$ (rationals) and $y=0$ (irrationals).